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University  of  California  •  Berkeley 

THE  THEODORE  P.  HILL  COLLECTION 

of 
EARLY  AMERICAN  MATHEMATICS  BOOKS 


DAY    AND    THOMSON'S    SERIES.      ' 

II  I  G  n  E  E 

ARITHMETIC; 


OB  THB 


SCIENCE  AND  APPLICATION  OF   NUMBERS; 


COMBINING  THB 


ANALYTIC  AXD  SYNTHETIC  MODES  OF  IXSTPOJCTION. 


DESIGNED   FOB 


ADVANCED  CLASSES  IN  SCHOOLS  AND  ACADEMIES. 


BY  JAMES  B.  THOMSON,  LL.D., 

OB  or  MENTAL  ARITHMETIC;  EXERCISES  m  ARITHMETICAL  ANALTBIS, 
PRACTICAL  ABITIIMKTIO;  EDITOR  OF  DAY'fl  SCHOOL  ALGEJSRA; 
LBGCNDRB'B  OEOMETBT,  ETC. 


ONE  HUNDRED  AND  TWENTIETH  EDITION. 


NEW    YORK: 

IYISON,  PHINNEY  &  CO,  48  &  50  WALKER  STREET. 

CHICAGO:  S.  C.  GRIGGS  &  CO.,  39  &  41  LAKE  ST. 
BOSTON:  BROWN,  TAQGARD  A  CHASE.     PHILADELPHIA:  SOWEK,  BARNES  *  oon 

AND   J.    B.    LIPPINCOTT    &    CO.      CINCINNATI;    MOORR,  WIL8TACII,    KEYS  A  CO. 

SAVANNAH:  j.  M.  COOPER  &  co.    BT.  Loms:  KEITH  &  WOODS.    NEW 
ORLEANS:  BLOOAIFIEXJ),  BTEEL  &  co.  DETROIT :>.  RAYMOND  A  co. 

1862. 


DAY  AND  THOMSON'S  MATHEMATICAL  SERIES. 

FOR  SCHOOLS  AND  ACADEMIES. 


I.  MENTAL  ARITHMETIC;  or,  First  Lessons  in  Numbers?— for  Begin. 
»ers.     This  work  commences  with  the  simplest,  combinations  of  numbers,  and 
gradually  advances  to  more  difficult  combinations,  as  the  mind  of  the  learnci 
expands  and  is  prepared  to  comprehend  them. 

II.  PRACTICAL    ARITHMETIC;  — Uniting  the   Inductive  with    the 
Synthetic  mode  of  Instruction;    also  illustrating  the  principles  ,of  CANCEI.A- 
Tiox.     The  design  of  this  work  is  to  make  the  pupil  thoroughly  acquainted 
with   the  reason  of  every  operation  which  he  is  required  to  perform.      It 
abounds  in  examples,  and  is  eminently  practical- 

III.  KEY  TO  PRACTICAL  ARITHMETIC ;— Containing  the  answers, 
with  numerous  suggestions,  &c. 

IV.  HIGHER  ARITHMETIC;  or,  the  Science  and  Application  of  Num- 
bers;— For  advanced  Classes.     This  work  is  complete  in  itself,  commencing 
with  the  fundamental  rules,  and  extending  to  the  highest  department  of  the 
science. 

V.  KEY  TO  HIGHER  ARITHMETIC;— Containing  all  the  answers, 
with  many  suggestions,  and  the  solution  of  the  more  difficult  questions. 

VI.  THOMSON'S  DAY'S  ALGEBRA;— This  work  is  designed  to  be  a 
Mcid  and  easy  transition  from  the  study  of  Arithmetic  to  the  higher  branches 
of  Mathematics.     The  number  of  examples  is  much  increased  ;  and  the  work 
is  every  way  adapted  to  the  improved  methods  of  instruction  in  Schools  and 
Academies. 

VII.  KEY    TO   THOMSON'S    DAY'S    ALGEBRA;— Containing  the 
answers,  the  solution  of  the  more  difficult  problems,  &c. 

VIII.  THOMSON'S    LEGENDRE'S    GEOMETRY ;— With    practical 
notes  and  illustrations.      This  work   has  received  the  approbation  of  many 
of  the  most  eminent  Teachers  and  Practical  Educators. 

IX.  PLANE    TRIGONOMETRY,   AND  THE  MENSURATION  OF 
HEIGHTS  AND  DISTANCES;  with  a  summary  view  of  the  Mature,  and 
Use  of  Logarithms ; — Adapted  to  the  method  of  instruction  in  Schools  and 
Academies 

X.  ELEMENTS  OF  SURVEYING ;— Adapted  both  to  the  wants  of  th 
(earner  and  the  practical  Surveyor. 


Entered  according  to  Act  of  Congress,  in  the  year  1847, 

BY  JAMES  B    THOMSON, 
In  the  Clerk's  Office  of  the  Northern  District  of  New  York. 


PREFACE. 


THE  Higher  Arithmetic  which  is  now  presented  to  the  public,  is  tho 
third  and  last  of  a  series  of  Arithmetics  adapted  to  the  wants  of  different 
classes  of  pupils  in  Schools  and  Academies.  The  title  of  each  explains 
the  character  of  the  work.  The  series  is  constructed  upon  the  principle, 
fhat  "  there  is  a  place  for  everything,  and  everything  should  be  in  its 
proper  place."  Each  work  forms  an  entire  treatise  in  itself ;  the  examples 
in  each  are  all  different  from  those  in  the  others,  so  that  pupils  who  study 
the  series,  will  not  be  obliged  to  purchase  the  same  matter  twice,  nor  to 
solve  the  same  problems  over  again. 

The  Mental  Arithmetic,  is  designed  for  children  from  six  to  eight  years 
of  age.  It  is  divided  into  progressive  lessons  of  convenient  length,  begin- 
ning with  the  simplest  combinations  of  numbers,  and  advancing  by  grad- 
ual steps,  to  more  difficult  operations,  as  the  mind  of  the  learner  expands 
and  is  prepared  to  comprehend  them. 

The  Practical  Arithmetic  embraces  all  the  subjects  requisite  for  a 
thorough  business  education.  The  principles  and  rules  are  carefully 
analyzed  and  demonstrated ;  the  examples  for  practice  are  numerous,  and 
the  observations  and  notes  contain  much  information  pertaining  to  busi- 
ness matters,  not  found  in  other  works  of  the  kind.  This  is  the  FIRST 
SCHOOL  BOOK  in  which  the  Standard  Units  of  Weights  and  Measures 
adopted  by  the  Government  in  1834,  were  published. 

The  Higher  Arithmetic  is  designed  to  give  a  full  development  of  the 
philosophy  of  Arithmetic,  and  its  various  applications  to  commercial  pur- 
poses. Its  plan  is  the  following : 

1.  The  work  is  complete  in  itself.  It  commences  with  notation,  and 
illustrating  the  different  properties  of  numbers,  the  principles  of  Cancela- 
tion,  and  various  other  methods  of  contraction,  extends  to  the  higher 
operations  in  mercantile  affairs,  and  the  more  abstruse  departments  of 
the  science.- 

%.  Great  pains  have  been  taken  to  render  the  definitions  and  rules  clear, 
concise,  exact,  comprehensive. 

3.  It  has  been  a  cardinal  point  never  to  anticipate  a  principle ;  and  never 
to  use  one  principle  in  the  explanation  of  another,  until  it  has  itself  been 
explained  or  demonstrated. 

4.  Nothing  is  taken  for  granted  which  requires  proof.     Every  principle 
therefore  has  been  retnvestigated,  and  carefully  analyzed. 


VI  PREFACE. 

5.  The  principles  are  at  ranged  consecutively  and  the  dependence  of 
each  on  those  that  precede  it,  is  pointed  out  by  references.     Treated  in 
this  manner,  the  science  of  Arithmetic  presents  a  series  of  principles  and 
propositions  alike  harmonious  and  logical ;  and  the  study  of  it  cannot  fail 
to  exert  the  happiest  influence  in  developing  and  strengthening  the  reason- 
ing powers  of  the  learner. 

6.  The  rules  are  demonstrated  with  care,  and  the  reasons  of  every  oper 
&tion  fully  illustrated. 

7.  The  examples  are  copious  and  diversified;  calling  every  principle 
into  exercise,  and  making  its  application  thoroughly  understood. 

8.  In  the  arrangement  of  subjects,  the  natural  order  of  the  science  ha* 
been  carefully  followed.     Common  Fractions  have  therefore  been  placed 
immediately  after  Division,  for  two  reasons.     First,  they  arise  from  divi- 
sion, and  a  connexion  so  intimate  should  not  be  severed  without  cause. 
Second,  in  Reduction  and  the  Compound  Rules,  it  is  often  necessary  to 
multiply  and  divide  by  fractions,  to  add  and  subtract  them,  also  to  carry 
for  them,  unless  perchance  the  examples  are  constructed  for  the  occasion 
and  with  special  reference  to  avoiding  these  difficulties. 

For  the  same  reason  Federal  Money,  which  is  based  upon  the  decimal 
notation,  is  placed  after  Decimal  Fractions ;  Interest,  Commission,  &c., 
after  Percentage.  To  require  a  pupil  to  understand  a  rule  before  he  is 
acquainted  with  the  principles  upon  which  it  is  based,  is  compelling  him 
to  raise  a  superstructure,  before  he  is  permitted  to  lay  a  foundation. 

9.  In  preparing  the  Tables  of  Weights  and  Measures,  no  effort  has 
been  spared  to  ascertain  those  in  present  use  in  our  country  ;  and  reject- 
ing such  as  are  obsolete,  we  havo  introduced  the  Standard  Units  adopted 
by  the  Government,  together  with  the  methods  of  determining  and  apply- 
ing those  standards. 

10.  Great  labor  has  also  been  expended  in  preparing  full  and  accurate 
Tables  of  Foreign  Weights  and  Measures,  and  Moneys  of  Account,  and 
in  comparing  them  with  those  of  the  United  States. 

Such  is  a  brief  outline  of  the  present  work.  In  a  word,  it  is  designed 
to  be  an  auxiliary  to  the  teacher,  a  lucid  and  comprehensive  text-book  for 
the  pupil,  and  an  acceptable  acquisition  to  the  counting-room.  It  contains 
many  illustrations  and  principles  not  found  in  other  works  before  the 
public,  and  much  is  believed  to  be  gained  in  the  method  of  reasoning 
and  analysis.  No  labor  has  been  spared  to  render  it  worthy  of  the 
marked  favor  with  which  the  former  productions  of  the  author  have 
been  received. 

J.  B.  THOMSON- 

NewY<rrk,  August,  1847. 


SUGGESTIONS 

ON  THE 

MODE  OF  TEACHING   ARITHMETIC. 


I.  QUALIFICATIONS/ — The  chief  qualifications  requisite  in  teaching  Arith- 
metic, as  well  as  other  branches,  are  the  following  : 

1.  A  thorough  knowledge  of  the  subject. 

2.  A  love  for  the  employment. 

3.  An  aptitude  to  teach.     These  are  indispensable  to  success. 

II.  CLASSIFICATION. — Arithmetic,  like  reading,  grammar,  &c.,  should  be 
taught  in  classes. 

1.  This  method  saves  much  time,  and  thus  enables  the  teacher  to  devote 
more  attention  to  oral  illustrations. 

2.  The  action  of  mind  upon  mind,  is  a  powerful  stimulant  to  exertion,  and 
iannot  fail  to  create  a  zest  for  the  study. 

3.  The  mode  of  analyzing  and  reasoning  of  one  scholar,  will  often  suggest 
new  ideas  to  others  in  the  class. 

4.  In  the  classification,  those  should  be  put  together  who  possess  as  nearly 
equal  capacities  and  attainments  as  possible.     If  any  of  the  class  learn  quicker 
than  others,  they  should  be  allowed  to  take  up  an  extra  study,  or  be  furnished 
with  additional  examples  to  solve,  so  that  the  whole  class  may  advance  together. 

5.  The  number  in  a  class,  if  practicable,  should  not  be  less  than  six,  nor 
over  twelve  or  fifteen.     If  the  number  is  less,  the  recitation  is  apt  to  be  defi- 
cient in  animation ;  if  greater,  the  turn  to  recite  does  not  come  round  suffi- 
ciently often  to  keep  up  the  interest. 

III.  APPARATUS. — The  Black-board  and  Numerical  Frame  areas  indispen- 
sable to  the  teacher,  as  tables  and  cutlery  are  to  the  house-keeper.     Not  a  reci- 
tation passes  without  use  for  the  black-board.     If  a  principle  is  to  be  demon- 
strated or  an  operation  explained,  it  should  be  done  upon  the  black-board,  so 
that  all  may  see  and  understand  it  at  once. 

To  illustrate  the  increase  of  numbers,  the  process  of  adding,  subtracting, 
multiplying,  dividing,  &c.,  to  young  scholars,  the  Numerical  Frame  furnishes 
one  of  the  most  simple  and  convenient  methods  ever  invented. 

Every  one  who  ciphers  will  of  course  have  a  slate.  Indeed,  it  is  desirable 
that  every  scholar  in  school,  even  to  the  very  youngest,  should  be  furnished 
with  a  slate,  so  that  when  their  lessons  are  learned  each  one  may  busy  himself 
In  writing  and  drawing  various  familiar  objects.  Idleness  in  school  is  the  parent 
tfriischirf,  and  employment  is  the  best  antidote  against  disobedience. 

Geometrical  diagrams  and  solids  are  also  highly  useful  in  illustrating  many 
points  in  arithmetic,  and  no  school  should  be  without  them. 

IV.  RECITATIONS. — The  first  object  in  a  recitation,  is  to  secure  the  attention 
of  the.  class.     This  is  done  chiefly  by  throwing  life  <ind  variety  into  Ihe  eier* 
citse.     Children  loaf-he  dullness,  while  animation  am    variety  are  iheii  delight,. 


Vlll  SUGGESTIONS. 

2.  Every  example  should  be  analyzed ;  the  "  why  and  the  wherefore"  of 
every  step  in  the  solution  should  be  required,  till  each  member  of  the  class  be- 
comes perfectly  familiar  with  the  process  of  reasoning  and  analysis. 

3.  To  ascertain  whether  each  pupil  has  the  right  answer,  it  is  an  excellent 
method  to  name  a  question,  then  call  upon  some  one  to  give  the  answer,  and 
before  deciding  whether  it  is  right  or  wrong,  ask  how  many  in  the  class  agree 
with  it.     The  answer  they  give  by  raising  their  hand,  will  show  at  once  how 
many  are  right.     The  explanation  of  the  process  may  now  be  made. 

V.  OBJECTS  OF  THE  STUDY. — When  properly  studied,  two  important  ends  are 
attained.    1st.  Discipline  of  mind,  and  the  development  of  the  reasoning  powers. 
2d.  Facility  and  accuracy  in  the  application  of  numbers  to  business  calculations. 

VI.  THOROUGHNESS. — The  motto  of  every  teacher  should  be  tJwroughness. 
Without  it,  the  great  ends  of  the  study  of  Arithmetic  are  defeated. 

1.  In  securing  this  object,  much  advantage  is  derived  from  frequent  reviews. 

2.  Every  operation  should  be  proved.     The  intellectual  discipline  and  habits 
of  accuracy  thus  secured,  will  richly  reward  the  student  for  his  time  and  toil. 

3.  Not  a  recitation  should  pass  without  practical  exercises  upon  the  black- 
board or  slates,  besides  the  lesson  assigned. 

4.  After  the  class  have  solved  the  examples  under  a  rule,  each  one  should 
be  required  to  give  an  accurate  account  of  its  principles  with  the  reason  for  each 
step,  either  in  his  own  language  or  that  of  the  author. 

5.  Mental  Exercises  in  arithmetic  are  exceeding Ly  useful  in  making  ready 
and  accurate  arithmeticians ;  hence,  the  practice  of  connecting  mental  with. 
written  exercises,  throughout  the  whole  course,  is  strongly  recommended. 

VII.  SELF-RELIANCE. — The  habit  of  self-reliance  in  study,  is  confessedly  in- 
valuable.    Its  power  is  proverbial;  I  had  almost  said,  omnipotent.     "  Where 
there  is  a  will,  there  is  a  way." 

1.  To  acquire  this  habit,  the  pupil,  like  a  child  learning  to  walk,  must  be 
taught  to  depend  upon  himself.     Hence, 

2.  When  assistance  is  required,  it  should  be  given  indirectly ;  not  by  taking 
the  slate  and  solving  the  example  for  him,  but  by  explaining  the  'meaning  of 
the  question,  or  illustrating  the  principle  on  which  the  operation  depends,  by 
supposing  a  more  familiar  case.      Thus  the  pupil  will  be  able  to  solve  the 
question  himself,  and  his  eye  will  sparkle  with  the  consciousness  of  victory. 

3.  The  pupil  should  be  encouraged  to  study  out  different  solutions,  and  to 
adopt  the  most  concise  and  elegant. 

4.  Finally,  he  should  learn  to  perform  examples  independent  of  the  answer. 
Without,  this  attainment  the  pupil  receives  but  little  or  no  discipline  from  the 
study,  and  acquires  no  confidence  in  his  own  abilities.     Wrhat  though  he  cornea 
to  the  recitation  with  an  occasional  wrong  answer;  it  were  better  to  solve  me 
question  unde/'slaiidingly  and  alone,  than  to  copy  a  score  of  answers  from  the 
book.     What  would  the  study  of  mental  arithmetic  be  worth,  if  the  pupil  had 
the  answers  before  him  1     What  is  a  young  man  good  for  in  the  counting-room, 
who  cannot  perform  arithmetical  operations  without  looking  to  the 

Every  on  3  pronounces  him  unjil  to  be  tiusted  with  bus  (lies?  calculations. 


CONTENTS. 


FAOt 

INTRID  JCTON, — a  brief  survey  of  the  science  of  Mathematics,         •  13 

SECTION   I. 
NOTJ  noN,  ........20 

Numeration,              -                                                -           -            «  24 

Formation  of  different  systems  of  Notation,        ...  i*J 

SECTION    II. 

GENERAL  RULE  for  Addition,                         .            .            .            .  3;} 

Diflenent  methods  of  proving  Addition,                -            -            -  3-1 

Counting-room  Exercises,      -...«••  37 

Adding  nuiul>crs  expressed  by  the  Roman  Nutation,      -            -  40 

SECTION    111. 

GENERAL  RCLE  for  Subtraction,        -            -            -            -            -  41 

Different  methods  of  proving  Subtraction,          -            -            .  -        4i 

Subtracting  numbers  expressed  liy  the  Roman  Notation,  47 

SECTION    IV. 

GENERAL  RULE  for  Multiplication,          -            -            -            -  -53 

Different  methods  of  proving  Multiplication,             ...  54 

Contractions  in  Multiplication,  six  methods,       -             -             -  5(3 

SECTION    V. 

GENERAL  RULE  for  Division,              -             -            .            -            -  71 

Different  methods  of  Proving  Division,                -            -            -  -        74 

Contractions  in  Division,  nine  methods,         -             -                           -  74 

General  principles  in  Division,                -             -             -             -  HI 

CA.NCEI.ATIHN,                                                     ....  8:5 

Applications  of  the  fundamental  rules,                 •             -           •-  b5 

SECTION    VI. 

PROPERTIES  OF  NUMBERS,                                          ...  80 

Method  of  finding  the  prime  numbers  in  any  series,                    -  -         J>3 

Table  of  prime  numbers  from  I  to  3-4 1 3.         -                       •    -  {•  1 

Numbers  changed  from  the  decimal  to  other  scales  of  Notation,  -         !<.") 

Analysis  of  Composite  Numbers,       .....  J)7 
Greatest  Common  Divisor,  two  methods,             ....       ](}(\ 

Least  Common  Multiple,  three  methods,        -                          -            -  102 

SECTION    VII. 

FRACTIONS,         -  ...       10* 

General  principles  pertaining  to  Fractions,                -            -            <•  K>3 

Reduction  of  Fractions,                -             -             -             -             -  .11 

Addition  of  Fractions,                          -             -             -             "         «  "  '  '** 
Subtraction  of  Fractions,                                                      ...        |-J-J 

Multiplication  of  Fractions,  common  method,            ...  lx.»l 

Multiplication  of  Fractions  by  Cancelation,        -  -        130 

Uonlractions  in  Multiplication  of  Fractions,  four  methods,    -  131 — 132 

1* 


X  CONTENTS. 

PAOB 

Division  af  Fractions,  common  method,              -            -            -            -  l.'J3 

Division  of  Fractions  by  Cancelation,                         -            -            -  136 

Contractions  in  Division  of  Fractions,  three  methods,                  -             -  l^H 

Complex  Fractions  reduced  to  Simple  ones,               ...  139 

SECTION    VIII. 

COMPOUND  NUMBERS,                               .....  144 

Tables  of  Compound  Numbers,          .....  144 

The  Standard  for  Gold  and  Silver  Coin  of  the  United  States,                -  145 

The  Standard  UNIT  of  Weight  of  the  United  States,                          -  147 

Origin  of  Weights  and  Measures,                                        -             -             •  147 

The  Avoirdupois  Pound  of  the  United  States  and  Great  Britain,      -  148 

The  Standard  UNIT  of  Length  of  the  United  States,      -                          -  150 

"                 "                "         of  New  York  and  Great  Britain,     -  151 

The  Standard  UNIT  of  Liquid  Measure  of  the  United  States,     -            -  154 

"             "             of  Dry  Measure  of  the  United  States,  155 

Method  of  determining  whether  a  given  year  is  Leap  Year,       -  157 

French  Money.  Weights  and  Measures,        -                                       -  l(»l 

Foreign  Weights  and  Measures,  compared  with  United  States,              -  KJ3 

GKNKRAL  RULE  for  Reduction,            .....  166 

Applications  of  Reduction,                         -                           ...  169 

Cubic  Measure  reduced  to  Dry,  or  Liquid  Measure,  &c.,     -  172 

Longitude  reduced  to  Time,  &c.,             -            -            -             -            -  175 

Compound  Numbers  reduced  to  Fractions,                ...  176 

Fractional  Compound  Numbers  reduced  to  whole  numbers,      -            -  179 

Compound  Addition,               ......  181 

Compound  Subtraction,               ......  183 

Comi.ouml  Multiplication,     ------  186 

Ccir.pound  Division,       -  -  -  -  -  -  -189 

SECTION   IX. 

DECIMAL  FRACTIONS,  their  origin,  &c.,         -            ...  191 

Method  of  reading  Decimals,       -                          -                                       -  193 

Addition  of  Decimals,              .____.  195 

Subtraction  of  Decimals,             ___._.  197 

Multiplication  of  Decimals,                 .....  199 

Contractions  in  Multiplication  of  Decimals,  two  cases,                -  201 

Division  of  Decimals,                                                     -                          -  205 

Contractions  in  Division  of  Decimals,                 ....  208 

Decimals  reduced  to  Common  Fractions,      -            -            -  210 

Common  Fractions  reduced  to  Decimals,            -                         -  2il 

Compound  Numbers  reduced  to  Decimals,                  ...  215 

SECTION   X. 

ADDITION  of  Circulating  Decimals,                 ....  222 

Subtraction  of  Circulating  Decimals,      .....  223 

MultJulication  of  Circulating  Decimals,         -                                     -  224 

Divis«/m  of  Cijculating  Decimals,           -            ...»  225 


CONTENTS.  XI 

PA8B 

SECTION  XI. 

FEDERAI   MONEY,      -                        .....  226 

Addition  of  Federal  Money,        -                                     ...  229 

Subtraction  of  Federal  Money,                       »  231 

Multiplication  of  Federal  Money,           .....  232 

Division  of  Federal  Money,                                          ...  234 

Counting-room  Exercises,  Contractions,  &c.      ....  237 

To  find  the  cost  of  articles  bought  and  sold  by  the  100  or  1000,        -  238 

Bills,  Accounts,  &c.,                                                          ...  239 

SECTION   XII. 

PERCENTAGE,  Percentage  Table,  &c.,            ....  241 

Applications  of  Percentage,        --....  244 

Commission,  Brokerage,  and  Stocks,             ....  244 

Commission  deducted  in  advance,  and  the  balance  invested,     -            -  2-17 

INTEREST,       -                                                  ....  2-19 

General  method  for  computing  Interest,              ....  252 

Second  method,  multiplying  the  Prin.  by  the  Int.  of  $1  for  the  time,  255 

To  compute  Interest  by  half  the  number  of  months,       ...  257 

To  compute  Interest  by  the  number  of  days,             ...  258 

Applications  of  Interest,  remarks  on  Promissory  Notes,  &c.,     -            -  258 

Forms  of  Negotiable  Notes,  &c.,      .....  2GO 

Partial  Payments  j  United  States  Rule,             -            -            -            -  2G1 

Connecticut  Rule,     -----..  263 

Vermont  Rule,                .......  2G4 

Problems  in  Interest,             ......  2G5 

Compound  Interest,        .......  270 

Discount,       -            -            -            -            -            -            -.  272 

Bank  Discount,               -                        .....  274 

To  find  what  sum  must  be  discounted  to  produce  a  given  amount,  277 

Insurance,  four  Cases,           ...                         _  278 

The  sum  to  be  insured  to  recover  the  premium  and  the  property,          -  282 

Life  Insurance,          .......  282 

Profit  and  Loss,  four  Cases,        ......  283 

Duties,  Specific  and  Advalorem,        .....  289 

Assessment  of  Taxes,      -.-....  292 

To  find  what  sum  must  be  assessed  to  raise  a  given  net  amount,     -  295 

Formation  of  Tax  Bills,                          .....  295 

Rate  Bills  for  Schools,                                                           -           -  297 

SECTION  XIII. 

ANALYSIS,                        -                                                ...  298 

Analyse  so.utions  of  questions  in  Simple  Proportion,                        -  299 

"                '•                "             Barter,           -            -            -            -  301 

"                "                "             Partnership,         ...  jju'J 

«               "               "            General  Average,      -           -           -  303 

«               «               «            Alligation,            ...  304 


XI  CONTENTS. 

Analytic  solutions  »>f  questions  in  Compound  Proportion,  -  307 

"  "  "  Position,  ...  308 

"  "  "  Practice,        -  -  •  309 

SECTION  XIV. 

RATIO,  general  principles  pertaining  to  it,  -  -  -  313 

Siirple  Proportion,  -  -  -  -  «  -  321 

Siirple  Proportion  and  its  Proof  by  Cancelation,      ...  325 

Compound  Proportion,  -  -   .          -  -  -  328 

Compound  Proportion  and  its  Proof  by  Cancelation,  -  -  330 

Conjoined  Proportion,     -  .....  332 

SECTION  XV. 

DUODECIMALS,  its  principles,  &c.,     .....  334 

Multiplication  of  Duodecimals,  .....  334 

SECTION  XVI. 

EQUATION  OP  PAYMENTS,     -  ....  ^38     ^ 

Partnership,        ........  340 

General  Average,       ----..-  343 

Exchange  of  Currencies,  ......  345 

Foreign  Coins  and  Moneys  of  Account,        ....  348 

Exchange,  Form  of  Bills  of  Exchange,  &c.,      -  -  -  -  351 

Arbitration  of  Exchange,       ......  355 

Alligation,  Medial,  and  Alternate,          ....  356 

SECTION  XVII. 

INVOLUTION,  and  Evolution,  -  -  ...  .  3GO 

Properties  of  Squares  and  Cubes,  .....  305 

Extraction  of  the  Square  Root,         .....  3»;7 

Demonstration  of  the  Square  Root,        .....  308 

Applications  of  the  Square  Root,       .....  370 

Extraction  of  the  Cube  Root,  Homer's  Method,  ...  374 

Demonstration  of  the  Cube  Root,     .....  376 

Extraction  of  Higher  Roots,       ---...  378 

SECTION  XVIII. 

Arithmetical  Progression,      ......  381 

Geometrical  Progression,  ......  384 

Annuities,          ----....  3^6 

Permutations  and  Combinations,  .....  3#g 

SECTION  XIX. 

Application  of  Arithmetic  to  Geometry,         ....  389 

Mensuration  of  Surfaces  and  Solids,      .....  3S9 

Measurement  of  timber,         --....  392 

Gauging  of  Casks,  -  -  .  .  .  .  .  393 

Tonnage  of  Vessels,  -»»••..  393 

Mechanical  Powers,        .......  394 

Miscellaneous  Examples,     -  .  .  .  .  .  395 


INTRODUCTION. 


ART,  1  •  Anything  which  can  be  multiplied,  divided ,  or  measutvd, 
is  called  QUANTITY.  Thus,  lines,  weight,  time,  number,  &c.,  are 
quantities. 

OBS.  1.  A  line  is  a  quantity,  because  it  can  be  measured  in  feet  and  inches; 
weight  can  be  measured  in  pounds  and  ounces;  time,  in  hours  and  minutes; 
numbers  can  be  multiplied,  divided,  &c. 

2.  Color,  and  the  operations  of  the  mind,  as  love,  hatred,  desire,  choice,  &c., 
cannot  be  multiplied,  divided,  or  measured,  and  therefore  cannot  properly  be 
called  quantities. 

2.  MATHEMATICS  is  the  science  of  Quantity. 

3*  The  fundamental  branches  of  Mathematics  are,  Arithmetict 
Algebra,  and  Geometry. 

4.  Arithmetic  is  the  science  of  Numbers. 

5  •  Alyebra  is  a  general  method  of  solving  problems,  and  of 
investigating  the  relations  of  quantities  by  means  cf  letters  and 
signs. 

OBS.  Fluxions,  or  the  Differential  and  Integral  Calculus,  may  be  considered 
as  belonging  to  the  higher  branches  of  Algebra. 

6.  Geometry  is  that  branch  of  Mathematics  which  treats  of 
Magnitude. 

7  •  The  term  magnitude  signifies  that  which  is  extended,  or 
which  has  one  or  more  of  the  three  dimensions,  length,  breadth,  and 
tiiickness,  Thus,  lines,  surfaces,  and  solids  are  magnitudes. 

QUEST.— 1.  What  is  Quantity  7    Give  some  examples  of  quantity.     O/ 1.  Why  is  a  lina 

*k  quantity  1    Weight  1    Time  ?    Numbers  ?    Are  color  and  the  operations  of  the  mind 

quantities  1     Why  not?    2.  What  is  Mathematics  1     3.   What  are  the  fundamental 

branches  of  mathematics  1   4.  What  is  Arithmetic  1   5.  Algebra  ?   6.  Geometry  1  1.  What 

!•  meant  by  magnitude  1 


14  INTRODUCTION. 

OBS.  1.  A  line  is  a  magnitude,  because  it  can  le  extended  in  length;  a 
surface,  because  it  has  length  and  breadth ;  a  solid,  because  it  has  length, 
breadth,  and  thickness. 

•I.  Motion,  though  a  quantity,  is  not,  strictly  speaking,  a  magnitude ;  for  it 
has  neither  length,  breadth,  nor  thickness. 

3.  The  term  magnitude  is  sometimes,  though  inaccurately,  used  as  synony- 
mous with  quantity. 

8  •  Trigonometry  and  Conic  Sections  are  branches  of  Mathemat- 
cs,  in  which  the  principles  of  Geometry  are  applied  to  triangles, 
and  the  sections  of  a  cone. 

9,  Mathematics  are  either  pure  or  mixed. 

In  pure  mathematics,  quantities  are  considered,  independently 
of  any  substances  actually  existing. 

In  mixed  mathematics,  the  relations  of  quantities  are  investi- 
gated in  connection  with  some  of  the  properties  of  matter,  or 
with  reference  to  the  common  transactions  of  business.  Thus,  in 
Surveying,  mathematical  principles  are  applied  to  the  measuring 
of  land  ;  hi  Optics,  to  the  properties  of  light ;  and  in  Astronomy, 
to  the  heavenly  bodies. 

OBS.  The  science  of  pure  mathematics  has  long  been  distinguished  for  the 
clearness  and  distinctness  of  its  principles,  and  the  irresistible  conviction  which 
they  carry  to  the  mind  of  every  one  who  is  once  made  acquainted  with  them. 
This  is  to  be  ascribed  partly  to  the  nature  of  the  subjects,  and  partly  to  the 
exactness  of  the  definitions,  the  axioms,  and  the  demonstrations. 

1 0.  A  definition  is  an  explanation  of  what  is  meant  by  a  word, 
or  phrase. 

OBS.  It  is  essential  to  a  complete  definition,  that  it  perfectly  distinguish*  the 
thing  defined,  from  everything  else. 

1 1  •  A  proposition  is  something  proposed  to  be  proved,  or 
required  to  be  done,  and  is  either  a  Theorem.,  or  a  Problem. 

1 2.  A  theorem  is  something  to  be  proved. 

1 3  •  A  problem  is  something  to  be  done,  as  a  question  to  be 
solved. 

(It  EST. — Obs.  Why  is  a  line  n  magnitude  ?  A  surface  ?  A  solid  7  Is  motion  a  magni- 
tude ?  Why  not?  &.  Of  how  many  kinds  are  mathematics  ?  In  pure  mathematics  how 
are  quantities  considered?  How  in  mixed  mathematics?  Obs.  For  what  is  the  science 
of  pure  mathematics  distinguished?  10.  What  is  a  definition  ?  Obs.  What  is  essential 
to  a  complete  definition  ?  11.  What  is  a  proposition  ?  12.  V  theorem ?  13.  A  problem? < 


INTRODUCTION.  15 

OBS.  1.  In  tie  statement  of  every  proposition,  whether  theorem  or  problem, 
certain  things  must  be  given,  or  assumed  to  be  true.  These  things  are  called 
the  data  of  the  proposition. 

2.  The  operation  by  which  the  answer  of  a  problem  is  found,  is  called  a 
solution. 

3.  When  tb.3  given  problem  is  so  easy,  as  to  be  obvious  to  every  one  without 
explanation,  i:  is  called  a  postulate. 

14.  One  proposition  is  contrary,  or  contradictory  to  another, 
when  what  is  affirmed  in  the  one,  is  denied  in  the  other. 

OBS.  A  proposition  and  its  contrary,  can  never  both  be  true.  It  cannct  be 
true,  that  two  given  lines  are  equal,  and  that  they  are  not  equal,  at  the  same 
time. 

1  5  •  One  proposition  is  the  converse  of  another,  when  the  order 
is  inverted ;  so  that,  what  is  given  or  supposed  in  the  first,  be- 
comes the  conclusion  in  the  last ;  and  what  is  given  in  the  last,  is 
the  conclusion,  in  the  first.  Thus,  it  can  be  proved,  first,  that  if 
the  sides  of  a  triangle  are  equal,  the  angles  are  equal ;  and  sec- 
ondly, that  if  the  angles  are  equal,  the  sides  are  equal.  Here,  in 
the  first  proposition,  the  equality  of  the  sides  is  given,  and  the 
equality  of  the  angles  inferred  ;  in  the  second,  the  equality  of  the 
angles  is  given,  and  the  equality  of  the  sides  inferred. 

OBS.  In  many  instances,  a  proposition  and  its  converse  are  both  true,  as  in 
the  preceding  example.  But  this  is  not  always  the  case.  A  circle  is  a  figure 
bounded  by  a  curve;  but  a  figure  bounded  by  a  curve  is  not  necessarily  a 
circle. 

16.  The  process  of  reasoning  by  which  a  proposition  is  shown 
to  be  true,  is  called  a  demonstration. 

OBS.  A  demonstration  is  either  direct  or  indirect. 

A  direct,  demonstration  commences  with  certain  principles  or  data  which  are 
admitted,  or  have  been  proved  to  be  true ;  and  from  these,  a  series  of  other 
truths  are  deduced,  each  depending  on  the  preceding,  till  we  arrive  at  the  truth 
which  was  required  to  be  established. 

An  indirect  demonstration  is  the  mode  of  establishing  the  truth  of  a  propo- 
sition by  proving  that  the  supposition  of  its  contrary,  involves  an  absurdity. 

QPEST. — Obs.  What  is  meant  by  the  data  of  a  proposition?    By  the  solation  cf 
problem  ?    What  is  a  postulate  1     14.   When   is  one  proposition  contrary  to  another 
Obs.  Can  a  proposition  and  its  contrary  both  be  true?    15.  When  is  one  proposition  the 
converse  of  another  ?     Obs.  Can  a  proposition  and  its  converse  both  be  true  ?     16.  What 
Is  a  demonstration  ?     Obs.  Of  how  many  kinds  are  demonstrations?    What  is  a  direct 
demonstration  ?     An  indirect  demonstration  ? 


16  INTRODUCTION. 

This  is  commonly  called  redudio  ad  absurdum.  The  former  is  the  more  com- 
mon method  of  conducting  a  demonstrative  argument,  and  is  the  meet  satisfac 
tory  to  the  mind.  » 

1 7  •  A  Lemma  is  a  subsidiary  truth,  or  proposition,  demon- 
strated for  the  purpose  of  using  it  in  the  demonstration  of  a 
theorem,  or  the  solution  of  a  problem. 

1 8.  A  Corollary  is  an  inference  or  principle  deduced  from  a 
preceding  proposition. 

1 0«  A  ScJiolium  is  a  remark  made  upon  a  preceding  prop- 
osition, pointing  out  its  connection,  use,  restriction,  or  extension. 

20.  An  Hypothesis  is  a  supposition,  made  either  in  the  state- 
ment of  a  proposition,  or  in  the  course  of  a  demonstration. 

AXIOMS. 

21.  An  Axiom  is  a  self-evident  proposition  ;  that  is,  a  prop- 
osition whose  truth  is  so  evident  at  sight,  that  no  process  of 
reasoning  can  make  it  plainer.     The  following  axioms  are  among 
the  most  common : 

1.  Quantities  wliich  are  equal  to  the  same  quantity,  are  equal 
to  each  other. 

2.  If  the  same  or  equal  quantities  are  added  to  equals,  the 
sums  will  be  equal. 

3.  If  the  same  or  equal  quantities  are  subtracted  from  equals, 
the  remainders  will  be  equal. 

4.  If  the  same  or  equal  quantities  are  added  to  unequals,  the 
sums  will  be  unequal. 

5.  If  the  same  or  equal  quantities  are  subtracted  from  unequals, 
the  remainders  will  be  unequal. 

6.  If  equal  quantities  are  multiplied  by  the  same  or  equal 
quantities,  the  products  will  be  equal. 

7.  If  equal  quantities  are  divided  by  the  same  or  equal  quan- 
tities, the  quotients  will  be  equal. 

8.  If  the  same  quantity  is  both  added  to  and  subtracted  from 
another,  the  value  of  the  latter  will  not  be  altered. 

QCTKST—  17.   What  is  a  lemma?    18.  What  is  a  corollary?    i9.  Wha-  is  a  scholium  1 
SO.  What  Is  an  hypothesis?    21.  What  is  an  axiom  1    Name  some  of  the  most  common 
not* 


INTRODUCTION.  17 

9.  If  a  quantity  is  both  multiplied  and  divided  by  the  same  or 
an  equal  quantity,  its  value  will  not  be  altered. 

10.  The  whole  of  a  quantity  is  greater  than  a  part. 

11.  The  whole  of  a  quantity  is  equal  to  the  sum  of  all  its  parts. 

SIGNS. 

22.  Addition  is  represented  by  the  sign  (-{-),  which  is  called 
plus.     It  consists  of  two  lines,  one  horizontal,  the  other  perpen- 
licular,  forming  a  cross,  and  shows  that  the  numbers  between 
which  it  is  placed,  are  to  be  added  together.    Thus,  the  expression 
6+8,  signifies  that  6  is  to  be  added  to  8.     It  is  read,  "  G  plus  8,"- 
or  "  0  added  to  8." 

OBS. — The  term  plus  is  a  Latin  word,  originally  signifying  "  more."  hence 
"added  to." 

23.  Subtraction  is  represented  by  a  short  horizontal  line  ( — ), 
which  is  called  minus.     When  placed  between  two  numbers,  it 
shows  that  the  number  after  it  is  to  be  subtracted  from  the  one 
before  it.     Thus,  the  expression  9 — 4,  signifies  that  4  is  to  bo 
subtracted  from  9 ;  and  is  read,  "  9  minus  4,"  or  "  9  less  4." 

OBS. — The  term  minus  is  a  Latin  word,  signifying  kst. 

24.  Multiplication  is  usually  denoted  by  two  oblique  lines 
crossing  each  other  (x),  called  the  sign  of  multiplication.     It 
shows  that  the  numbers  between  which  it  is  placed,  are  to  be 
multiplied  together.     Thus,  the  expression  (9x0),  signifies  that 
9  and  C  are  to  be  multiplied  together,  and  is  read,  "  9  multiplied 
by  6,"  or  simply,  "9  into  0."     Sometimes  multiplication  is  de- 
noted by  a  point  (.)  placed  between  the  two  numbers  or  quanti- 
ties.    Thus,  9.6  denotes  the  same  as  9X6. 

OBS.  It  is  Letter  to  denote  the  multiplication  of  figures  by  a  cross  than  by  a 
point ;  for  the  latter  is  liable  to  be  confounded  with  the  decimal  point. 

24.  a.  When  two  or  more  numbers  are  to  be  subjected  to  the 
same  operation,  they  must  be  connected  by  a  line  ( )  placed 


show  1  Obs.  What  is  the  meaning  of  the  term  phis  7  23.  How  is  subtraction  represented  1 
What  is  the  sign  of  subtraction  called  1  What  does  it  show?  Obs.  What  does  the  term 
minus  signify  7  24.  How  is  multiplication  usually  denoted  ?  What  does  the  sign  of  mul- 
tiplication show  7  In  what  other  way  is  multiplication  sometimes  denoted  1 


18  INTRODUCTION. 

over  them,  called  a  vinculum,  or  by  a  parenthesis  (  ).  Thus  the 
expression  (12-{-3)x2,  shows  that  the  sum  of  12  and  3,  is  to  be 
multiplied  by  2,  and  is  equal  to  30.  But  12-J-3X2,  signifies 
that  3  only  is  to  be  multiplied  by  2,  and  that  the  product  is  to 
be  added  to  12,  which  will  make  18. 

25.  Division  is  expressed  in  two  ways  : 

First,  by  a  horizontal  line  between  two  dots  (-5-),  called  the 
sign  of  division,  which  shows  that  the  number  before  it,  is  to  be 
divided  by  the  number  after  it.  Thus,  the  expression  24—6 
signifies  that  24  is  to  be  divided  by  6. 

Second,  division  is  often  expressed  by  placing  the  divisor  under 
the  dividend,  in  the  form  of  a  fraction.  Thus,  the  expression 
3j£,  shows  that  35  is  to  be  divided  by  7,  and  is  equivalent  to 
35V7. 

26.  The  equality  between  two  numbers  or  quantities,  is  rep- 
resented by  two  parallel  lines  (=),  called  the  sign  of  equality. 
Thus,  the  expression  5-|-3=8,  denotes  that  5  added  to  3  are 
equal  to  8.     It  is  read,  "  5  plus  3  equal  8,"  or  "  the  sum  of  5 
plus  3  is  equal  to  8."     So  7+5=16 — 4=12. 


QTTEST.— 24.  a.  When  two  or  more  numbers  are  to  be  subjected  to  the  same  operation, 
what  must  be  done  ?  25.  In  how  many  ways  is  division  expressed  1  What  is  the  first  ? 
What  does  this  sign  show  ?  What  is  the  second  ?  26.  How  is  the  equality  between  two 
numbers  or  quantities  represented? 


ARITHMETIC, 

SECTION    I. 
NOTATION  AND  NUMERATION. 

ART.  27,  Any  single  thing,  as  a  peach,  a  rose,  a  book,  is 
called  a  unit,  or  one  ;  if  another  single  thing  is  put  with  it,  the 
collection  is  called  two  ;  if  another  still,  it  is  called  three  ;  if  an- 
other, four  ;  if  another,  Jive,  &c. 

The  terms,  one,  two,  thtw,  tec.,  by  which  we  express  how  rwny 
single  things  or  units  are  under  consideration,  are  the  names  of 
numbers.  Hence, 

28,  NUMBER  signifies  a  unit,  or  a  collection  of  units. 

OES.  1.  Numbers  are  divided  into  two  classes,  abstract  and  concrete. 

When  they  are  applied  to  particular  objects,  as  two  pears,  Jive  pounds,  ten 
dollars,  &c.,  they  are  called  concrete  numbers. 

When  they  do  not  refer  to  any  particular  object,  as  when  we  say  four  and 
five  are  nine,  they  are  called  abstract  numbers. 

2.  Whole  numbers  are  often  called  integers. 

3.  Numbers  have  various  properties  and  relations,  and  are  applied  to  variou» 
computations  in  the  practical  concerns  of  life.     These  properties  and  applica- 
tions are  formed  into  a  system,  called  Arithmetic. 

29.  ARITHMETIC  is  the  science  of  numbers. 

OBS.  1.  The  term  Arithmetic  is  derived  from  the  Greek  word  arithmgiiMt, 
which  signifies  the  art  of  reckoning  by  numbers. 

2.  The  aid  of  Arithmetic  is  required  to  make  and  apply  calculations  not 
only  in  faisiness  transactions,  but  in  tJmost  every  department  of  mathematics. 


.  What  is  a  single  thing  called?  If  another  is  put  with  it,  what  is  the  col- 
lection called  1  If  another,  what?  What  are  the  terms  one,  two,  three,  &c.  ?  28.  What 
does  number  signify?  Obs.  Into  how  many  classes  are  numbers  divided?  When  are 
they  called  concrete?  When  abstract?  To  what  are  numbers  applied  1  29.  What  to 
Arithmetic  1  Obs.  In  what  is  the  aid  of  arithmetic  require  i  ? 
T.H.  2 


20  NOTATION.  [SECT.  I 

Numbers  are  expressed  by  words,  by  letters,  and  by  figures. 

NOTATION. 

30.  The  art  of  expressing  numbers  by  letters  or  figures,  ii 
called  NOTATION.     There  are  two  methods  of  notation  in  use,  the 
Roman  and  the  Arabic. 

3 1 .  The  Roman  method  employs  seven  capital  letters,  viz :  I, 
V,  X,  L,  C,  D,  M.     When  standing  alone,  the  letter  I,  denotes 
one  ;  V,  Jive  ;  X,   ten  ;  L,  fifty  ;  C,   one  hundred  ;  D,  five  hun- 
dred ;  M,  one  thousand.    To  express  the  intervening  numbers  from 
one  to  a  thousand,  or  any  number  larger  than  a  thousand,  we  re- 
sort to  repetitions  and  various  combinations  of  these  letters.     The 
method  of  doing  this  will  be  easily  learned  from  the  following 

TABLE. 


I      denotes 

one. 

XXX  denote 

thirty. 

II 

ft 

two. 

XL 

forty. 

III 

tt 

three. 

L 

fifty. 

IV 

it 

four. 

LX 

sixty. 

V 

tt 

five. 

LXX       « 

seventy. 

VI 

tt 

six. 

LXXX    « 

eighty. 

VII 

tt 

seven. 

XC 

ninety. 

VIII 

ft 

eight. 

C 

one  hundred. 

IX 

tt 

nine. 

CI 

one  hundred  and 

me. 

X 

« 

ten. 

CX 

one  hundred  and 

t.en. 

XI 

« 

eleven. 

cc 

two  hundred. 

XII 

tt 

twelve. 

ccc 

three  hundred 

XIII 

ft 

thirteen. 

COCO     " 

four  hundred. 

XIV 

tt 

fourteen. 

D 

five  hundred. 

XV 

tt 

fifteen. 

DC 

six  hundred. 

XVI 

ft 

sixteen. 

DCC       " 

seven  hundred. 

XVII 

tt 

seventeen. 

DCCC     " 

eight  hundred. 

XVIII 

ft 

eighteen. 

DCCCC  " 

nine  hundred. 

XIX 

tt 

nineteen. 

M 

one  thousand. 

XX 

ft 

twenty. 

MM 

two  thousand. 

XXI 

tt 

twenty-one. 

MDCCCLV, 

one  thousand  e 

lgh« 

XXII 

tt 

twenty-two,  &c. 

hundred 

and.  fifty-five.    . 

.  —  How  are  numbers  usually  expressed  1     30.  What  is  notation? 
methods  are  there  in  use  1    31.  What  is  employed  by  the  Roman  method  ? 


How  nwm> 


ARTS.  30-33.]  NOTATION.  21 

OBS.  1.  This  method  of  expiessing  numbers  was  invented  by  the  Romans, 
and  is  therefore  called  the  Roman  Notation.  It  is  now  seldom  used,  except  to 
denote  chapters,  sections,  and  other  divisions  of  books  and  discourses. 

2.  The  letters  C  and  M,  are  the  initials  of  the  Latin  words  centum^  and 
mille,  the  former  of  which  signifies  a  hundred,  and  the  latter  a  thousand:  for 
this  reason  it  is  supposed  they  were  adopted  to  represent  these  numbers. 

3 1  •  a.  It  will  be  perceived  from  the  Table  above,  that  every 
time  a  letter  is  repeated,  its  value  is  repeated.  Thus  I,  standing 
alone,  denotes  one  ;  II,  two  ones,  or  two,  &c.  So  X  denotes  ten ; 
KX,  twenty,  &c. 

When  a  letter  of  a  less  value  is  placed  before  a  letter  of  a 
greater  value,  the  less  takes  away  its  own  value  from  the  greater ; 
but  when  placed  after,  it  adds  its  own  value  to  the  greater. 

32.  A  line  or  bar  ( — )   placed  over  a  letter,  increases  its 
value  a  thousand  times.     Thus,  V  denotes  five,  V  denotes  five 
thousand ;  X,  ten  ;  X,  ten  thousand,  &c. 

OBS.  1.  In  the  early  periods  of  this  notation,  four  was  written  IIII,  instead 
of  IV ;  nine  was  written  VIIII,  instead  of  IX ;  forty  was  written  XXXX, 
instead  of  XL,  &c. 

The  former  method  is  more  convenient  in  performing  arithmetical  operations 
In  addition  and  subtraction ;  while  the  latter  is  shorter  and  better  adapted  to 
ordinary  purposes. 

2.  A  thousand  was  originally  written   CIO,  which,  in  later  times,  was 
changed  into  M ;  Jive  hundred  was  written  10  instead  of  D.     Annexing  O  to 
10  increased  its  value  ten  times.     Thus,  IOO  denoted  Jive  thousand;  IOOO, 
fifty  thousand,  &c. 

3.  Prefixing  C  and  annexing  O  to  the  expression  CIO,  makes  its  value  ten 
times  greater:  thus,  CCIOO  denotes  ten  thousand-,    CCCIOOO,  a  hundred 
thousand.     According  to  Pliny,  the  Romans  carried  this  mode  of  notation  no 
further.     When  they  had  occasion  to  express  a  larger  number,  they  did  it  by 
repetition.   Thus,  CCCIOOO,  CCCIOOO,  expressed  two  hundred  thousand,  &c. 

33.  The  common  method  of  expressing  numbers  is  by  the 
Arabic  Notation.     The  Arabic  method  employs  the  following  ten 
cJtaracters  or  figures,  viz : 

12         34567         '8  90 

one,  two,  three,  four,  five,  six,  seven,  eight,  nine,  zero. 


QrKST.~O6s.  Why  is  this  method  called  Roman  ?  31.  a.  What  is  the  effect  of  repeating 
a  letter?  If  a  letter  of  less  value  is  placed  before  another  of  greater  value,  what  is  th» 
effect?  If  placed  after,  what?  32.  When  a  line  or  bar  is  placed  over  a  letter,  how  does 
it  affect  its  value  ?  33.  What  is  the  common  way  of  expressing  numbers  1  How  many 
characters  does  this  method  employ  1 


22  NOTATION.  [SECT.  1 

The  first  nine  are  called  significant  figures,  because  each  one 
always  has  a  value,  or  denotes  some  number.  They  are  also 
called  digits,  from  the  Latin  word  digitus,  which  signifies  a 
finger. 

The  last  one  is  called  a  cipher,  or  naught,  because  when  stand- 
ing alone  it  has  no  value,  or  signifies  nothing. 

OBS.  1.  It  must  not  be  inferred,  however,  that  the  cipher  is  useless;  for  when 
placed  on  the  right  of  any  of  the  significant  figures,  it  increases  their  value. 
It  may  therefore  be  regarded  as  an  auxiliary  digit,  whose  office,  it  will  be  seen 
hereafter,  is  as  important  as  that  of  any  other  figure  in  the  system. 

2.  Formerly  all  the  Arabic  characters  were  indiscriminately  called  ciphers ; 
hence  the  process  of  calculating  by  them  was  called  ciphering ;  on  the  same 
principle  that  calculating  by  figures  is  called  figuring. 

34.  It  will  be  seen  that  nine  is  the  greatest  number  that  can 
be  expressed  by  any  single  figure  in  the  Arabic  system  of  Nota- 
tion. 

All  numbers  larger  than  nine  are  expressed  by  combining  to- 
gether two  or  more  of  these  ten  figures,  and  assigning  different 
values  to  them,  according  as  they  occupy  different  places.  For 
example,  ten  is  expressed  by  combining  the  1  and  0,  thus  10 ; 
eleven  by  two  Is,  thus  11 ;  twelve  by  1  and  2,  thus  12 ;  twenty, 
thus  20  ;  thirty,  thus  30  ;  &c.  A  hundred  is  expressed  by  com- 
bining the  1  and  two  Os,  thus  100;  two  hundred,  thus  200;  a 
thousand  by  combining  the  1  and  three  Os,  thus  1000;  two  thou- 
sand, thus  2000  ;  ten  thousand,  thus  10,000  ;  a  hundred  thousand, 
thus  100,000;  a  million,  thus  1,000,000;  tea  millions,  thus 
10,000,000;  &c.  Hence, 

35.  The  digits  1,  2,  3,  <fcc.,  standing  alone,  or  in  the  right 
hand  place,  respectively  denote  units  or  ones,  and  are  called  units 
of  the  Jirst  order.  * 

When  they  stand  in  the  second  place,  they  express  tens,  or  ten 
ones  ;  that  is,  their  value  is  ten  times  as  much  as  when  standing 

QUEST.— What  are  the  first  nine  called  1  Why  ?  What  else  are  they  called  ?  What 
fe  the  last  one  called?  Why?  Obs.  Is  the  ciphe.  useless?  What  may  it  be  regarded  ? 
What  is  the  origin  of  the  term  ciphering?  34.  What  is  the  greatest  number  that  can  be 
expressed  by  one  figure  1  How  are  larger  numbers  expressed?  35.  What  do  the  digits, 
i,  2,  3,  &c.,  denote,  when  standlig  alone,  or  in  the  right  hand  place?  What  are  they 
then  called?  What  do  they  den j to  when  standing  in  the  second  plaoe) 


ARTS.  34-37.J  NOTATION  2.3 

in  the  first  or  rigl.it  hand  place,  and  they  are  called  units  of  the 
second  order. 

When  occupying  the  third  place,  they  express  hundreds  ;  that 
is,  their  value  is  ten  times  as  much  as  when  standing  in  the  sec- 
ond place,  and  they  are  called  units  of  the  third  order. 

When  occupying  the  fourth  place,  they  express  thousands  ;  that 
is,  then*  value  is  ten  times  as  much  as  when  standing  in  the  third 
place,  and  they  are  called  units  of  the  fourth  order,  &c.  Thus,  it 
will  be  seen  that, 

Ten  units  make  one  ten,  ten  tens  make  one  hundred,  and  ten  hun 
dreds  make  one  thousand  ;  that  is,  ten  in  an  inferior  order  are  equal 
to  one  in  the  next  superior  order.  Hence,  universally, 

36.  Numbers  increase  from  right  to  left  in  a  tenfold  ratio; 
consequently  each  removal  of  a  figure  one  place  towards  the  left,  in- 
creases its  value  ten  times. 


.  —  1.  The  number  which  forms  the  basis,  or  which  expresses  the  ratio 
of  increase  in  a  system  of  Notation,  is  called  the  RADIX  of  that  system.  Thus, 
the  radix  of  the  Arabic  notation  is  ten. 

2.  The  reason  that  numbers  increase  from  right  to  left,  instead  of  left  to 
right,  is  probably  owing  to  the  ancient  practice  of  writing  from  the  right  hand 
to  the  left. 

37.  The  different  values  which  the  same  figures  have,  are  called 
simple  and  local  values. 

The  simple  value  of  a  figure  is  the  value  which  it  expresses 
when  it  stands  alone,  or  in  the  right  hand  place.  Hence  the  sim- 
ple value  of  a  figure  is  the  number  which  its  name  denotes. 

The  local  value  of  a  figure  is  the  increased  value  which  it  ex- 
presses by  having  other  figures  placed  on  its  right.  Hence  the 
local  value  of  a  figure  depends  on  its  locality,  or  the  place  which 

QUEST.—  What  Is  their  value  then  ?  What  are  they  called  1  What  Is  a  figure  called 
when  it  occupies  the  third  place  7  What  is  its  value  then  ?  What  is  it  called  when  in 
the  fourth  place  ?  What  is  its  value  1  How  many  units  are  required  to  make  one  ten  ? 
How  many  tens  make  a  hundred  ?  How  many  hundreds  make  a  thousand  1  How  many 
of  an  inferior  order  are  required  to  make  one  of  the  next  superior  order  ?  36.  What  is  the 
general  law  by  which  numbers  increase  ?  What  is  the  effect  upon  the  value  of  a  figure 
to  remove  it  one  place  towards  the  left  ?  Note.  What  is  the  number  called  which  forms 
the  basis  jr  the  ratio  of  increase  in  a  system  of  notation  1  What  is  the  radix  of  the  Arable 
notation?  Why  do  numbers  increase  from  right  to  left?  37.  What  are  the  different 
values  of  the  same  figure  called  1  What  Is  the  simple  value  of  a  figure  ?  What  the  local  1 


24  NUMERATION.  [SECT.  I. 

it  occupies  in  relation,  to  other  numbers  with  which  it  is  connected. 
(Art.  35.) 

OBS.  1.  This  system  of  notation  is  called  Arabic,  because  it  is  supposed  to 
tave  been  invented  by  the  Arabs. 

2.  It  is  also  called  the  decimal  system,  because  numbers  increase  in  a  ten- 
fold ratio.     The  term  decimal  is  derived  from  the  Latin  word  decem,  which  sig- 
nifies ten. 

3.  The  early  history  of  the  Arabic  notation  is  veiled  in  obscurity.     It  is  th* 
opinion  of  some  whose  judgment  is  entitled  to  respect,  that  it  was  invented  by 
the  philosophers  of  India.     It  was  introduced  into  Europe  from  Araoia  abou 
Jhs  eighth  century,  and  about  the  eleventh  century  it  came  into  general  use, 
xrfii  in  England  and  on  the  continent.     The  application  of  the  term  digit  to 
the  significant  figures,  affords  strong  presumptive  evidence  that  the  system  had 
its  origin  in  the  ancient  mode  of  counting  and  reckoning  by  means  of  the 
fingers ;  and  that  the  idea  of  employing  ten  characters,  instead  of  twelve  or 
any  other  number,  was  suggested  by  the  number  of  fingers  and  thumbs  on  both 
hands.    (Art.  33.) 

NUMERATION. 

38.  The  art  of  reading  numbers  when  expressed  by  figures,  tt 
called  NUMERATION. 

The  pupil  will  easily  learn  to  read  the  largest  numbers  from  the 
following  scheme,  called  the 


NUMERATION   TABLE. 


I 


685,  876,  389,  764,  391,  827,  218,  649,  853,  123,  234,  579,  793,  465,  623. 
"xvt  xrv!  xin.  xii.    xi.    x.    ix.  vin.  vn.   vi.     v.    iv.    in.     n.     i. 

39.  The  different  orders  of  numbers  are  divided  into  periods 
of  taree  figures  each,  beginning  at  the  right  hand.  The  first 
period,  which  is  occupied  by  units,  tens,  hundreds,  is  called  uniti 

Q.UI8T. — Upon  what  does  the  local  value  of  a  figure  depend  1  Obs.  Why  is  this  system 
of  notation  called  Arabic  ?  What  else  is  it  sometimes  called  1  Why  ?  What  do  you 
•ay  of  its  early  history?  When  was  it  introduced  into  Europe  ?  What  is  the  probable 
origin  of  the  system  ?  Why  were  ten  characters,  rather  than  any  other  number,  adopted  7 
*&  What  is  Numeration  1  39.  How  are  the  orders  of  numbers  divided  1  Wha  is  the 
§1*1  period  called  1  By  what  is  it  occupied  1 


ARTS.  38, 39.]  NUMERATION.  25 

period  ;  the  second  is  occupied  by  thousands,  tens  of  thousands, 
hundreds  of  thousands,  and  is  called  thousands'  period ;  the  third 
is  occupied  by  millions,  tens  of  millions,  hundreds  of  millions,  and 
is  called  millions'  period  ;  the  fourth  is  occupied  by  billions,  tens 
of  billions,  hundreds  of  billions,  and  is  called  billions'  period  ;  and 
so  on,  the  orders  of  each  successive  period  being  units-,  tens,  and 
hundreds. 

The  figures  in  the  table  are  read  thus:  685  tredecillions,  876 
duodecillions,  389  undecillions,  764  decillions,  391  nonillions,  827 
octillions,  218  septillions,  649  sextillions,  853  quintillians,  123 
quadrillions,  234  trillions, -579  billions,  793  millions,  465  thou- 
sand, 6  hundred  and  twenty-three. 

Note. — 1.  The  terms  thirteen,  fourteen,  fifteen,  &c.,  are  obviously  derived 
from  three  and  ten,  four  and  ten,  five  and  ten,  which  by  contraction  become 
thirteen,  fourteen,  fifteen,  and  are  therefore  significant  of  the  numbers  which 
they  denote.  The  terms  eleven  and  twelve,  are  generally  regarded  as  primitive 
words ;  at  all  events,  there  is  no  perceptible  analogy  between  them  and  tho 
numbers  which  they  represent.  Had  the  terms  oneteen  and  twoteen  been 
adopted  in  their  stead,  the  names  would  then  have  been  significant  of  the 
numbers  one  and  ten,  two  and  ten;  and  their  etymology  would  have  been 
similar  to  that  of  the  succeeding  terms. 

The  terms  twenty,  thirty,  forty,  &c.,  were  formed  from  two  tens,  three  tens, 
four  tens,  which  were  contracted  into  twenty,  thirty,  forty,  &c. 

The  terms  twenty-one,  twenty-two,  twenty-three,  &c.,  are  compounded  of 
twenty  and  one,  twenty  and  two,  &c.  All  the  other  numbers  as  far  as  ninety- 
nine,  are  formed  in  a  similar  manner. 

2.  The  terms  hundred,  thousand  and  million  are  primitive  words,  and  bear 
uo  analogy  to  the  numbers  which  they  denote.    The  numbers  between  a  hun- 
dred and  a  thousand  are  expressed  by  a  repetition  of  the  numbers  below  a 
hundred.     Thus  we  say  one  hundred  and  one,  one  hundred  and  two,  one 
hundred  and  three,  &c. 

3.  The  terms  billion,  trillion,  quadrillion,  &c.,  are  formed  from  million  and 
the  Latin  numerals  bis,  tres,  quatuor,  &c.     Thus,  prefixing  bis  to  million,  by  a 
slight  contraction  for  the  sake  of  euphony,  it  becomes  billion ;  prefixing  tres  to 
million,  it  is  easily  contracted  into  trillion,  &c.     The  Latin  word  bis  signifies 
two ;  tres,  three ;  quatiwr,  four ;  quinque,  five ;  sex,  six ;  scptcm,  seven ;  octot 
eight ;  novem,  nine ;  decem,  ten ;  undecim,  eleven ;  duodecim,  twelve ;  tredecvm, 
thirteen. 

Q.UEST. — What  is  the  second  period  called  ?  By  what  occujiod?  What  is  the  third 
tailed  ?  By  what  occupied  1  What  is  the  fourfl  called  1  By  w  hat  occupied  ?  Wha  t  H 
he  fifth  called  1  By  what  occupied?  Repeat  the  Numeration  Table,  beginning  at  the 
«ht  hand. 

3 


26 


NUMERATION. 


(.SECT.  I. 


Higher  periods  than  those  in  the  Table,  may  be  easily  formed  by  following 
the  above  analogy. 

4.  The  foregoing  law,  which  assigns  superior  values  to  these  ten  characters, 
according  to  the  order  or  place  which  they  occupy  and  the  use  of  so  many 
derivative  and  compound  words  in  forming  the  names  of  numbers,  saves  an 
inconceivable  amount  of  time  and  labor  in  learning  Notation  and  Numeration, 
as  well  as  in  their  application. 

4.-O.  To  read  numbers  which  are  expressed  by  figures. 

Point  them  off  into  periods  of  three  figures  each  ;  tlien,  beginning 
t  the  left  Jiand,  read  the  figures  of  each  period  in  tlie  same  manner 
as  those  of  tlie  right  hand  period  are  read,  and  at  the  end  of  each 
period,  pronounce  its  name. 

OBS.  1.  The  learner  must  be  careful,  in  pointing  off  figures,  always  to  begin 
at  the  rigid  hand ;  and  in  reading  them,  to  begin  at  the  left  hand. 

2.  Since  the  figures  in  the  first  or  right  hand  period  always  denote  units, 
its  name  is  not  pronounced.  Hence,  in  reading  figures,  when  no  period  ia 
mentioned,  it  is  always  understood  to  be  the  right  hand,  or  units'  period.j 

EXERCISES   IN    NUMERATION. 


Note. — In  numerating  large  numbers,  it  is  advisable  for  the  pupil  first  to 
apply  to  each  figure  the  name  of  the  order  which  it  occupies.  Thus,  beginning 
at  the  right  hand,  he  should  say,  "  Units,  tens,  hundreds,"  &c.,  and  point  at 
the  same  time  to  the  figures  standing  in  the  order  which  he  mentions. 

Read  the  following  numbers : 


Ex.  1. 

3506 

11. 

706305 

21. 

967058713 

2. 

6034 

12. 

1640030 

22. 

32100040 

3. 

5060 

13. 

830006 

23. 

106320000 

4. 

90621 

14. 

70900038 

24. 

780507031 

5. 

73040 

15. 

3067300 

25. 

4063107 

6. 

450302 

16. 

12604321 

26. 

29038450 

7. 

603260 

17. 

70003000 

27. 

1046347025 

8. 

130070 

18. 

161010602 

28. 

20380720000 

9. 

2021305 

19. 

80367830 

29. 

8503467039 

10. 

4506580 

20. 

400031256 

30. 

450670412463 

QUEST. — 40.  How  do  you  read  numbers  expressed  by  figures  7  Cbs.  Where  begin  to 
point  them  off"?  Where  to  read  theml  Do  you  pronounce  the  natae  of  the  right  hand 
period  ?  When  no  period  is  named,  what  is  understood  ? 


ARTS.  40,  41.]  NUMERATION.  2*7 


31.  130812000641 

32.  5200240301000 

33.  98760000216 

34.  82600381000000 

35.  403070003462000 


36.  120340078910356 

37.  43601000345000 

38.  506302870045380 

39.  42008120537062035 

40.  653107843604893048 


41.  210  256  031  402  385  290  845  381  467. 

42.  361  438  201  219  763  281  572  829  318  278. 

4 1  •  The  method  of  dividing  numbers  into  periods  of  three  fig- 
res,  was  invented  by  the  French,  and  is  therefore  called  the 
French  Numeration. 

The  English  divide  numbers  into  periods  of  six  figures,  in  the 
following  manner : 


II  II 

0  s  °  a 

%  «  fg  s   .  .g 

S  *  a  *  g  <s  §  g  « 

s  ^   -  2    §  §  ^  I  I 

^  s  s  g  „-  s  s  5  S  < 


g  S  S  g          g  I  £  S  §  £ 

i        o  I  ;  o  i        ^  . . 

M  ^H^^S  ^H^ 


•sJS'si  'sJS'sl  *  J 

rfi      C_(     ^        75      i^A  rf>      t—>      r^        W      ^  ») 

S  *S    1  ?  "S    |  S  'S    g    S  "5    g  S^«2 

"§  g  1 1  S  -S  "§  s  I 'S  a  I  i  g  I  a  j  .-j 

hP  H  H  K  H  *"3  ^EHEH^nE'1^  K  E"1  H  ffi  E"1  *^ 

423561,  234826,  479365 


Period  III.  Period  II.  Period  I. 

According  to  this  method,  the  preceding  figures  are  read  thus  : 
423561  billions,  234826  millions,  and  479365. 

OBS.  1.  It  will  be  perceived  that  the  two  methods  agree  as  far  as  hundreds 
of  millions;  the  former  then  begins  a  new  period,  while  the  latter  continues  on 
through  thousands  of  millions,  &c. 

2.  The  French  method  is  generally  used  throughout  the  continent  of  Europe, 
as  well  as  in  America,  and  has  been  recently  adopted  by  some  English  authors. 
It  b  very  genei  illy  admitted  to  be  more  simple  and  convenient  than  the  Eng- 
lish method. 


.—  41.  What  is  the  French  method  of  numeration  1    \V~fat  the  English  method  f 
Ois.  Which  is  the  more  simple  and  convenient  1 

2* 


28  NOTATION.  [SECT.  1. 

EXERCISES  IN  NOTATION. 

4  2  •  To  express  numbers  by  figures. 

Begin  at  the  left  hand,  and  write  in  each  order  the  figure  which 
denotes  the  given  number  in  that  order. 

If  any  intervening  orders  are  omitted  in  the  proposed  number, 
write  ciphers  in  their  places.  (Art.  38.) 

Write  the  following  numbers  in  figures  : 

1.  Two  thousand,  one  hundred  and  nine. 

2.  Twenty  thousand  and  fifty-seven. 

3.  Fifty-five  thousand  and  three. 

4.  One  hundred  and  five  thousand,  and  ten. 

5.  Seven  hundred  and  ten  thousand,  three  hundred  and  one. 

6.  Two  millions,  sixty-three  thousand,  and  eight. 

7.  Fourteen  millions,  and  fifty-six. 

8.  Four  hundred  and  forty  millions,  and  seventy-two. 

9.  Six  billions,  six  millions,  six  thousand,  and  six. 

10.  Forty-five  billions,  three  hundred  and  forty  thousand,  and 


11.  Five  hundred  and   fifty-six  millions,  three  thousand,  two 
hundred  and  sixty-four. 

12.  Eight  hundred  and  ten  billions,  ten  millions,  and  seventy- 
five  thousand. 

13.  Ninety-six  trillions,  seven  hundred  billions,  and  fifty-four. 

14.  Three  hundred  and  forty-nine  quadrillions,  five  trillions, 
seven  billions,  four  millions,  and  twenty. 

15.  Nineteen  quintillions. 

16.  Six  hundred  and  thirty  sextillions. 

17.  Two  hundred  and  ninety-eight  septillions. 
18.»Seventy-four  octillions. 

19.  Four  hundred  and  ten  decillions. 

20.  Eight  hundred  and  sixty-three  duodecillions. 

21.  Nine  hundred  and  thirty-five  tredecillions. 

22.  Six  hundred  and  seventy-three  quintillions,  seventeen  quad 
i  vlions,  and  forty-five. 

23.  Twenty  trillions,  six  hundred  and  forty-eight  billions,  and 
t»renty-five  thousand. 


ARTS.  42-44. j  NOTATION.  29 

OBS.  The  great  facility  with  which  large  numbers  may  be  expressed  both 
in  language  and  by  figures,  is  calculated  to  give  an  imperfect  idea  of  their  real 
magnitude.  It  may  assist  the  learner  in  forming  a  just  conception  of  a  million, 
a  billion,  a  trillion,  &c.,  to  reflect,  that  to  count  a  million,  at  the  rate  of  a  hun- 
dred a  minute,  would  require  nearly  seventeen  days  often  hours  each;  to  count 
a  billion,  at  the  same  rate,  would  require  more  than  forty-Jive  years;  and  /</» 
count  a  trillion,  more  than  45,662  years. 

43.  From  the  preceding  illustrations,  the  learner  will  per- 
ceive that  a  variety  of  other  systems  of  notation  may  be  formed 
upon  the  same  principle,  having  different  numbers  for  their 
radices.  Thus,  if  we  wished  to  form  a  quinary  system ;  that  is, 
a  system  in  which  the  numbers  should  increase  in  &  five-fold  ratio, 
or  has  five  for  its  radix,  it  would  require  four  significant  figures 
and  a  cipher.  Let  the  figures  1,  2,  3,  4,  and  0,  be  the  characters 
employed ;  then  five  would  be  expressed  by  1  and  0,  and  would 
be  Avritten  thus  10 ;  six  by  1  and  1,  thus  11  ;  seven  by  1  and  2, 
thus  12  ;  eight  by  1  and  3,  thus  13  ;  nine  by  1  and  4,  thus  14; 
ten  by  2  and  0,  thus  20 ;  eleven  by  2  and  1,  thus  21,  &c. 

4:4.  In  the  binary  or  diadic  system  of  notation  developed  by 
Leibnitz,  there  are  two  characters  employed,  1  and  0.  The  cipher 
when  placed  at  the  right  hand  of  a  number,  in  this  system,  mul- 
tiplies it  by  two.  Thus  the  number  one  is  expressed  by  1 ;  two 
by  10;  three  by  11  ;  four  by  100;  five  by  101;  six  by  110; 
seven  by  111;  eight  by  1000;  nine  by  1001;  ten  by  1010; 
eleven  by  1011,  &c. 

OBS.  1.  In  like  manner  other  systems  of  notation  may  be  formed,  having 
three,  four,  six,  eight,  twelve,  or  any  given  number  for  their  radix. 

When  the  radix  is  two,  the  system  is  called  bi.nary  or  diadic ;  when  three 
it  is  called  ternary ;  when  four,  quaternary ;  when  five,  quinary ;  when  six, 
senary;  when  seven,  septenary;  when  eight,  odary;  when  nine,  nonary,  &c 

2.  It  should  be  observed  that  every  system  of  notation,  formed  upon  the 
foregoing  principles,  will  require  as  many  distinct  characters,  as  there  are  urJts 
In  the  radix,  and  that  one  of  them  must  be  a  cipher,  and  another  a  unit. 

For  the  method  of  changing  numbers  from  the  decimal  to  other  scales  of 
notation,  and  the  converse,  see  Arts.  162,  163. 

QUEST. — 43.  Is  the  decimal  notation  the  only  system  that  can  be  formed  on  the  same 
•Manciples!  How  would  you  form  a  quinary  system  of  notation?  Wiite  six  in  the  qui- 
»ary  scale  on  the  black-board.  Write  seven,  nine,  ten,  eleven,  twelve  Ob$.  How  many 
•haracters  will  any  system  formed  upon  this  principle  require  1 

3* 


30  NOTATION.  [SECT.  [. 

45.  About  the  commencement  of  the  second  century,  Ptolemy 
introduced  the  sexagesimal  notation,  which  has  sixty  for  its  radix. 

OBS.  1.  It  is  said  that  the  Chinese  and  some  other  eastern  nations  now  em- 
ploy this  system  in  measuring  time,  using  periods  of  sixties,  instead  of  centuries. 
Relics  of  the  sexagesimal  notation  may  also  be  seen  in  our  division  of  the  circle,, 
and  of  time,  where  the  degree  and  hour  are  each  divided  into  60  minutes,  the 
minute  into  (50  seconds,  &c. 

2.  The  Roman  notation  seems  to  have  been  commenced  with  V  or  five  for 
its  radix,  which  was  afterwards  changed  to  X  or  ten.  It  may  therefore  be 
regarded  as  a  kind  of  combination  of  the  quinary  and  decimal  systems. 

46.  Since  the  number  eight  may  be  divided  and  sub-divided 
so  many  times  without  a  remainder,  some  contend  that  a  system 
of  notation  having  eight  for  its  radix,  would  be  preferable  to  the 
decimal  system. 

Others  claim  that  the  duodecimal  notation ;  that  is,  a  system 
with  twelve  for  its  radix,  would  be  more  convenient  than  either." 
However  this  may  be,  the  decimal  system  is  so  firmly  rooted,  it 
were  hopeless  to  attempt  a  change. 

OBS.  It  may  be  doubted  whether  any  other  ratio  of  increase  would,  on  the 
whole,  be  more  convenient,  than  that  of  the  present  system.  If  the  ratio  were 
less,  it  would  require  more  places  of  figures  to  express  large  numbers ;  if  the  ratio 
were  larger,  it  would  not  indeed  require  so  many  figures,  but  the  operations, 
would  manifestly  be  more  difficult  than  at  present,  on  account  of  the  numbers 
in  each  order  being  larger.  Besides,  the  decimal  system  is  sufficiently  compre- 
hensive to  express  with  all  desirable  facility,  every  conceivable  number,  the 
largest  as  well  as  the  smallest ;  and  yet  it  is  so  simple,  that  a  child  may  under- 
stand and  apply  it.  In  a  word,  it  is  every  way  adapted  to  the  practical  ope- 
rations of  business,  as  well  as  the  most  abstruse  mathematical  investigations 
In  whatever  light,  therefore,  it  is  viewed,  the  decimal  notation  must  be  re- 
garded as  one  of  the  most  striking  monuments  of  human  ingenuity,  and  iti 
beneficial  influence  on  the  progress  of  science  and  the  arts,  on  commerce  and 
civilization,  must  win  for  its  unknown  author  the  everlasting  admiration  and 
gratitude  of  mankind. 

*  Barlow's  Theory  of  Numbers,  Leslie's  Plalosophy  of  Arithmetic,  Edi  r  CTji  Eaej 
elope  dia. 


ARTS.  45-50.]  ADDITION.  31 

SECTION    II. 

ADDITION. 

ART.  49.  Ex.  1.  A  man  bought  three  lots  of  land ;  the  first 
contained  23  acres,  the  second  9  acres,  and  the  third  15  acres : 
how  many  acres  did  he  buy  ? 

Solution. — 23  acres  and  9  acres  are  32  acres,  and  15  are  47 
aeres.  Ans.  47  acres. 

OBS.  It  will  be  seen,  that  the  solution  of  this  example  consists  in  finding  a 
tingle  number,  which  will  exactly  express  the  value  of  the  several  given  num- 
bers untied  together. 

5O.  The  process  of  uniting  two  or  more  numbers  together -,  so 
as  to  form  a  single  number,  is  called  ADDITION. 

The  answer,  or  the  number  thus  found,  is  called  the  Sum  or 
Amount. 

OBS.  When  the  numbers  to  be  added  are  all  of  the  same  denomination,  as 
all  dpllars,  all  pounds,  &c.,  the  operation  is  called  Simple  Addition. 

Ex.  2.  A  miller  bought  7864  bushels  of  wheat  of  one  man, 
4952  bushels  of  another,  and  3273  bushels  of  another :  how 
many  bushels  did  he  buy  of  all  ? 

Write  the  numbers  under  each  other,  so  that  Operation. 

units  may  stand  under  units,  tens  under  tens,  7864 

&c.,  and  draw  a  line  beneath  them.     Then  be-  4952 

ginning  at  the  right  hand  or  units,  add  each  3273 

column  separately.  Thus,  3  units  and  2  units  Ans.  16089  bu. 
are  5  units,  and  4  are  9  units.  Write  the  9  in  units'  place  under 
the  column  added.  Next  7  and  5  are  12,  and  6  are  18  tens.  But 
18  requires  two  figures  to  express  it ;  (Art.  34  ;)  consequently  it 
cannot  all  be  written  under  its  own  column.  We  therefore  write 
the  8  or  right  hand  figure  in  tens'  place  under  the  column  added, 
and  reserving  the  1  or  left  hand  figure,  add  it  with  the  hundreds. 
Thus,  1  which  was  reserved,  and  2  are  3,  and  9  are  12,  and  8  are 
20  hundreds.  Set  the  0  or  right  hand  figure  under  the  column 

QUEST.— 50.  What  is  Addition  •?  What  is  the  answer  called  1  Obs.  When  the  nuiu 
bers  to  be  added  are  all  of  the  same  denomination,  what  is  the  opera  don  called  1  51.  \Vha4 
order,!  of  figures  4oyou  add  together  ? 


32  ADDITION.  [SECT.  II. 

add«<i,  and  reserving  the  2  or  left  hand  figure,  add  it  to  the  next 
column  as  before.  Thus,  2  which  were  reserved  and  3  are  5, 
and  4  are  9,  and  7  are  16  thousands.  Set  the  6  under  the  col- 
umn added  ;  and  since  there  is  no  other  column  to  be  added, 
write  the  1  in  the  next  place  on  the  left. 

5 1  •  It  will  be  perceived  in  this  example,  that  units  are  added 
to  units,  tens  to  tens,  &c. ;  that  is,  figures  of  the  same  order  are 
added  to  each  other.  All  numbers  must  be  added  in  the  same 
manner.  For,  figures  standing  in  different  orders  or  columns  ex- 
press different  values  ;  (Art.  35  ;)  consequently,  they  cannot  be 
united  together  directly  in  a  single  sum.  Thus,  3  units  and  5 
tens  will  neither  make  eight  units,  nor  eight  tens,  any  more  than  3 
oranges  and  5  apples  will  make  8  apples,  or  8  oranges.  In  like 
manner  it  is  plain  that  7  tens  and  2  hundreds  will  neither  make 
9  tens,  nor  9  hundreds. 

OBS.  The  object  of  writing  units  under  units,  tens  under  tens,  &c.,  is  to 
prevent  mistakes  which  might  occur  from  adding  different  orders  to  each  other. 

52.  When  the  sum  of  a  column  does  not  exceed  9,  it  will  be 
noticed,  we  set  it  under  the  column  added ;  but  if  it  exceeds  9, 
we  set  the  units  or  right  Jiand  figure  under  the  column  added, 
and  reserving  the   tens  or  left  hand  figure,  add  it  to  the  next 
column.     In  adding  the  last  column  on  the  left,  we  set  down  the 
whole  sum. 

OBS.  The  process  of  reserving  tJie  tens,  or  left  hand  figure,  and  adding  it  to 
the  next  column,  is  called  carrying  tens. 

53.  The  principle  of  carrying  may  be  illustrated  in  the  follow- 
ing  manner. 

Take,  for  instance,  the  last  example,  7864 

and  adding  as  before,  write  the  sum  of  4952 

each  column  in  a  separate  line.     Thus,  3273 

the  sum  of  the  units'  column  is  9  units  ;  9  sum   of  units, 

the  sum  of  the  tens'  column  is  18  tens,  or  18*     "      "    tens. 

1  hundred  and  8   tens  ;  the  sum  of  the  19**     "      "  hund. 

hundreds'  column  is   19  hundred,   or  1  14***     "      "  thou. 

thousand   9   hundred;   the  sum   of  the  16089    Amount. 

Q.uicsT.«-Why  not  add  figures  of  different  orders  together  1 


ARTS.  51— 55. J  ADDITION.  33 

thousands'  column  is  14  thousand.  Now,  adding  these  results 
together  as  they  stand,  units  to  units,  tens  to  tens,  <fcc.,  the  amount 
is  16089  bushels,  which  is  the  same  as  in  the  solution  above. 

Thus,  it  is  evident,  when  the  sum  of  a  column  exceeds  9,  the 
right  hand  figure  denotes  units  of  the  same  order  as  the  column 
added,  and  the  tens  or  left  hand  figure  denotes  units  of  the  next 
higher  order.  Hence, 

The  reason  we  carry  the  tens  or  left  hand  figure  to  the  next 
column,  is  because  it  is  of  the  same  order  as  the  next  column,  and 
figures  of  the  same  order  must  always  be  added  togetlier.  (Art.  51.) 

OBS.  1.  The  reason  for  setting  down  the  whale  sum  of  the  last  or  left  hand 
column,  is  because  there  are  no  figures  in  the  next  order  to  which  the  left 
hand  figure  can  be  added.  It  is,  in  fact,  carrying  it  to  the  next  column. 

2.  From  the  preceding  illustration  it  will  also  be  seen,  that  the  object  of 
beginning  to  add  at  the  right  hand  is,  that  we  may  carry  t/ie  tens,  as  we  pro- 
ceed in  the  operation. 

54.  From  the  preceding  illustrations  and  principles  we  de- 
rive the  following 

GENERAL  RULE  FOR  ADDITION. 

I.  Write  the  numbers  to  be  added,  under  each  other  /  50  that 
units  may  stand  under  units,  tens  under  tens,  &c.    (Art.  51.  Obs.) 

II.  Begin  at  the  right  hand,  and  add  each  column  separately. 
When  the  sum  of  a  column  does  not  exceed  9,  write  it  under  the 
column  ;  but  if  the  sum  of  a  column  exceeds  9,  write  the  units' 
figure  under  the  column  added,  and  carry  the  tens  to  the  next 
column.  (Arts.  52,  53.) 

III.  Proceed  in  this  manner  through  all  the  orders,  and  set  down 
the  whole  sum  of  the  last  or  left  hand  column.     (Art.  53.    Obs.) 

55.  PROOF. — Beginning  at  the  top,  add  each  column  down- 
wards, and  if  tJie  second  result  is  the  same  as  the  first,  the  work  is 
supposed  to  be  right. 

GUEST.  54.— How  do  you  write  numbers  to  be  added  7  Why  place  units  under  units,  &c.  1 
Where  do  you  begin  to  add  ?  When  the  sum  of  a  column  does  no  exceed  9,  what  do  you 
do  with  it  ?  When  it  exceeds  9,  how  proceed  1  What  is  mea:»t  by  carrying  the  tens  1 
Why  carry  the  tens  to  the  next  column?  Why  begin  to  add  a  the  right  hand  ?  What 
do  you  do  with  the  sum  of  the  last  column  ?  55.  How  is  additio  t  proved  1 


34  ADDITION.  [SECT.  II. 

Note. — The  object  of  beginning  at  the  top  and  adding  downwards,  is  that 
the  figures  may  be  taken  in  a  different  order  from  that  in  which  they  were 
added  before.  The  order  being  reversed,  the  presumption  is,  that  any  mistake 
which  may  have  been  made  will  thus  be  detected ;  for  it  can  hardly  be  sup- 
posed that  two  mistakes  exactly  equal  will  occur. 

56,  Second  Method — Cut  off  the  bottom  line,  and  find  the 
sum  of  the  rest  of  the  numbers ;  then  add  this  sum  and  the  bot- 
tom line  together,  and  if  the  second  result  is  the  same  as  the  first, 
the  work  is  supposed  to  be  right. 

Note. — 1.  This  method  of  proof  depends  on  the  axiom,  that  the  whole  of  a 
quantity  is  equal  to  the  sum  of  all  its  parts.  (Ax.  11.) 

2.  The  method  of  cutting  off  the  top  line,  and  afterwards  adding  it  to  the 
sum  of  the  others,  is  objectionable  on  account  of  adding  the  numbers  in  the 
same  order  as  they  were  added  hi  the  solution.  (Art.  55.  Note.) 

5*7.  Third  Method. — From  the  amount,  subtract  all  the  given  numbers  but 
one,  and  if  the  remainder  is  equal  to  the  number  not  subtracted,  the  work  may 
be  supposed  to  be  right. 

Note. — This  method  supposes  the  pupil  to  be  acquainted  with  subtraction 
before  he  commences  this  work.  It  is  placed  here  on  account  of  the  con- 
venience of  having  all  the  methods  of  proving  the  rule  together. 

58.  Fourth  Method* — Cast  the  9s  out  of  each  of  the  given  numbers  sepa- 
rately, and  place  each  excess  at  the  right  of  the  number.  Then  cast  the  9a 
out  of  the  sum  of  these  excesses ;  also  cast  the  9s  out  of  the  amount ;  and  if 
these  two  excesses  are  equal,  the  work  may  be  supposed  to  be  right. 

Note. — 1.  This  mode  of  proof  is  based  on  a  peculiar  property  of  the  number  9. 
For  its  illustration  and  demonstration,  see  Art.  161.  Prop.  14. 

2.  To  cast  the  9s  out  of  a  number,  begin  at  the  left  hand,  add  the  digits, 
together,  and,  as  soon  as  the  sum  is  9  or  over,  drop  the  9,  and  add  the  remain- 
der to  the  next  digit,  and  so  on.  For  example,  to  cast  the  9s  out  of  4626357, 
we  proceed  thus :  4  and  6  are  10 ;  drop  the  9  and  add  the  1  to  the  next  figure. 
1  and  2  are  3,  and  6  are  9  ;  drop  the  9  as  above.  3  and  5  are  8,  and  7  are 
15  ;  dropping  the  9,  we  have  6  remainder. 

EXAMPLES    FOR   PRACTICE. 

59.  Ex.  1.  A  man  paid  2468  dollars  for  his  farm,  1645  dollars 
for  a  house,  865  dollars  for  stock,  and  467  dollars  for  tools;  how- 
much  did  he  pay  for  the  whole  ? 

2.  A  produce  merchant  bought  5  cargoes  of  corn ;  the  first  con- 

QTJEST.— Note.  Why  add  the  columns  downwards,  instead  of  upwards  ?  Can  addition 
be  proved  by  any  other  methods  ? 

*  Wallis'  Arithmetic,  Oxford  165T. 


ARTS.  56-59.]  ADDITION.  35 

tained  6725  bushels,  the  second  7208,  the  third  5047,  the  fourth 
12386,  and  the  fifth  10391  bushels:  how  many  bushels  did  he 
buy? 

3.  A  tavern-keeper  bought  six  loads  of  hay  which  weighed  as 
follows:  1725  pounds,  2163  pounds,  1581  pounds,  1908  pounds, 
2340  pounds,  and  1879   pounds:    what  was  the  weight  of  the 
whole  ? 

4.  A  man  gave  5460  dollars  to  his  oldest  son,  to  the  next  4065, 
o  the  next  6750,  to  the  next  8000,  and  to  the  youngest  7276 

dollars :  how  much  did  he  give  to  all  ? 

5.  A  merchant,  on  settling  up  his  business,  found  he  owed  one 
creditor  176  dollars,  another  841  dollars,  another  1356  dollars, 
another  2370  dollars,  another  840  dollars :  what  was  the  amount 
of  his  debts  ? 

6.  The  state  of  Maine  contains  32400  square  miles ;  New  Hamp- 
shire, 9500;  Vermont,  9700;  Massachusetts,  7800;  Rhode  Island, 
1251 ;  and  Connecticut,  4789  :  how  many  square  miles  are  there 
in  the  New  England  States  ? 

7.  The  state  of  New  York  contains  46220  square  miles ;  New- 
Jersey,  7948  ;  Pennsylvania,  46215  ;   and  Delaware,  2068  :   how 
many  square  miles  are  there  in  the  Middle  States  ? 

8.  The  state  of  Maryland  contains  10755  square  miles;  Virginia, 
65700;  North  Carolina,  51632  ;  South  Carolina,  31565  ;  Georgia, 
61683;   Florida,  56336;  Alabama,  54G$4 ;  Mississippi,  49356; 
Louisiana,  47413 ;  and  Texas,  100000 :   how  many  square  miles 
are  there  in  the  Southern  States  ? 

9.  The  state  of  Tennessee  contains  41752;  Kentucky,  40023; 
Ohio,  40500  ;  Michigan,  60537  ;  Indiana,  35626  ;  Illinois,  56506  ; 
Missouri,  70050:  Arkansas,  54617;  Iowa,  173786;  and  Wiscon- 
sin, 92930 ;   how  many  square  miles  are  there  in  the  Western 
States? 

10.  What  is  the  whole  number  of  square  miles  in  the  United 
States  ? 

11.  What  is  the  sum  of  75234+41015  +  19075+1764-88350 
+  10040? 

12.  What  is  the  sum  of  250120+30402+7850+465000  + 
10046+65045? 


36  ADDITION.  [SECT.  II. 

13.  What  is  the  sum  of  85046  +  90045+412260+125781  + 
4060  +  273048? 

14.  What  is  the  sum  of  1500267+45085+4652  +  4780400  + 
90276  +  89760841? 

15.  What   is   the   sum   of  45702125  +  67070420+670856  + 
4230825  +  750642  +  8790845  ? 

16.  What  is  the  sum  of  825760842  +  35620476  +  7800490  f 
467243  +  98371  +  6425  +  740  ? 

17.  What  is  the  sum  of  2503  +  37621+475290  +  1223729  + 
10671840  +  275600312? 

18.  What  is  the  sum  of  463270+2500  +  7200342  +  10271  + 
426345  +  6200705? 

19.  What  is  the  sum  of  80429  +  7562345  +  700100  +  85261798 
+4000101  +  3007002? 

20.  What  is  the  sum  of  756  +  849+934+680  +  720+843  + 
657689  +  989876498  +  8045685+807266780? 

21.  What  is  the  sum  of  6457  +  29301  +  82406  +  7589  +  63489 
+101364+46745? 

22.  Add  together  786,  840,  910,  403,  783,  650,  809,  670,  408, 
310,  and  652. 

23.  Add   together  16075,   250763,   7561,    830654,   293106, 
2537104,  and  316725. 

24.  Add  together  25^  40,  751,  302,  75,  831,  26,  43,  621,  340, 
and  510. 

25.  Add  together  493742,  56710607,  23461,  400072,  6811004, 
8999003,  and  26501. 

26.  Add  together  629405,  7629,  31000401,  263012,  1300512, 
390217,  and  13268. 

27.  Add  together  286013,  4016702,  1971342,  6894680,  28945, 
and  2624302. 

28.  Add  together  460167,  296345,  84634123,64205,  9673108, 
and  1931456. 

29.  Add  together  432678902,  310046734,  2167005,  327861 
asid  293000428. 


ART.  60.J 


ADDITION. 


«eOUNTING-ROOM   EXERCISES. 

GO,  To  the  accountant  as  well  as  the  mathematician,  accuracy 
and  expertness  in  adding,  are  indispensable.  These  attainments  can 
be  acquired  only  by  frequent  exercises  in  footing  up  long  columns 
of  figures. 

Note. — 1.  Instead  of  saying  4  and  8  are  12,  and  2  arc  14, 
and  7  are  21,  and  4  are  25,  &c.,  a  skilful  accountant,  per- 
forming the  addition  at  a  glance,  simply  pronounces  the 
results.  Thus,  four,  twelve,  twenty-one,  thirty-one,  (4-J-6 
=  10,)  thirty-seven,  forty-seven,  (7-f-3=;lO,)  fifty-two. 

2.  When  two  or  three  figures  taken  together  make  10,  as 
6  and  4,  or  2,  3,  and  5,  &c.,  it  accelerates  the  process  to 
add  their  sum  at  once.     A  little  practice  will  enable  the 
student  to  run  up  a  long  column  of  figures  with  as  much 
facility  almost  as  he  can  count. 

3.  When  the  columns  are  long,  accountants  sometimes 
set  the  figure  to  be  carried  below  the  other  figure  under  the 
column  added.     Thus,  the  sum  of  the  first  column  in  the 
example  above  being  52,  set  the  5  (the  figure  carried)  be- 

.ow  the  2.  The  sum  of  the  second  column  being  48,  set  the  4  below  the  8,  &c. 
This  method  saves  much  time  in  reviewing  an  operation,  and  also  enables  us, 
when  interrupted,  to  resume  the  process  where  we  left  off. 

Required  the  amount  of  each  of  the  following  e  samples : 

31. 

Dollars. 
2425 


30. 

86015 

251  03 

85057 

12236 

43026 

67084 

21167 

54042 

42158 

240.34 

459982 

Aiis. 

3045 

Car. 

3282 
2793 
2354 
4262 
9158 
2653 
3424 
12G6 
8742 
2126 
5387 

Ans.  47872 
Car.   465 


32. 

Dollars. 
46,519 
32,271 
17,436 
81,587 
28,333 
52,745 
23,052 
20,158 
71,232 
39,464 
18,643 
42,027 
73,235 
24,103 


33. 

Yards. 
607,253 
232,012 
211,849 
380,436 
578,551 
231,349 
145,763 
605,037 
760,155 
357,676 
544,844 
276,232 
803,383 
725,918 


34. 

Pounds. 
421,536 
310,101 
797,019 
233,680 
124,402 
255,353 
852,057 
618,041 
100,266 
971,134 
536,920 
703,352 
420,503 
312675 


3S  ADDITION.  [SECT.  II. 


85. 

36. 

37. 

38. 

318,037 

460,375 

963,172 

849,652 

272,465 

841,681 

300,725 

361,728 

530,634 

239,724 

463,248 

412,381 

109,871   4 

763,256 

721,003 

635,403 

693,036 

437,891 

387,356 

872,545 

764,543 

825,432 

241,653 

406,223 

323,638 

285,678 

603,280 

294,867 

428,432 

310,720 

532,176 

811,236 

389,763 

403,521 

278,321 

576,037 

210,045 

687,489 

829,248 

213,744 

760,806 

324,061 

171,320 

764,368 

636,215 

530,724 

206,782 

305,216 

253,734 

623,452 

461,027 

436,720 

251,600 

487,638 

589,203 

823,284 

575,453 

290,731 

248,639 

217,436 

807,720 

803,256 

730,461 

592,301 

930,046 

731,463 

672,398 

243,762 

174,173 

379,574 

246,175 

731,445 

626,245 

823,156 

928,340 

429,374 

342,734 

928,348 

731,629 

684,569 

61.  Accountants  often  acquire  the  habit  of  adding  two  col- 
umns of  figures  at  a  time.  The  power  of  rapid  addition  is  easily 
acquired,  and  is  well  worthy  the  attention  of  the  student.  The 
following  examples  will  illustrate  the  principle. 

39.  What  is  the  sum  of  312817+527236  +  141625+462415 
+  251818  +  234112? 

Operation. 

Taking  the  two  right  hand  columns,  we  312817 

say,  12  and  18  are  30,  and  15  are  4C    and  527236 

25  are  70,  and  36  are  106,  and  17  are  123.  141625 

Set  down  the  23  under  the  columns  added,  462415 

and  carry  the  1  or  left  hand  figure  to  the  251818 

column  of  hundreds.     Proceed  in  the  same  234112 

manner  with  the  other  columns.  Ans>  1930023 


ART.  61.] 


ADDITION. 


(41.)    (42.)    (43.) 


21 
30 
11 
13 
20 
15 
34 
18 
12 
17 
23 


22 
13 
40 
25 
14 
11 
33 
45 
12 
20 
18 


44 
20 
25 
17 
50 
14 
16 
28 
11 
14 
37 


(44.) 
1325 
1510 
1314 
3141 
1016 
2233 
1224 
2415 
1830 
1814 
1621 


(45.) 
2610 
1511 
1021 
1115 
1513 
4020 
1316 
1233 
2515 
1718 
2142 


(46.) 
344235 
402321 
141511 
201250 
154036 
132212 
181714 
213025 
111817 
161518 
432733 


What  was  the  amount  of  exports  and  imports  of  the  United 
States  in  1840,  and  of  shipping  in  1842  ? 


(47.) 

(48.) 

(49.) 

States. 

Exports. 

Imports. 

Shipping. 

Maine,      .         Dolls 

.  1,018,269 

Dolls.  628,762 

T.  281,930 

N.  Hampshire, 

20,979 

114,647 

23,921 

Vermont, 

305,150 

404,617 

4,343 

Massachusetts, 

10,186,261 

16,513,858 

494,895 

Rhode  Island, 

206,989 

274,534 

47,243 

Connecticut, 

518,210 

277,072 

67,749 

New  York, 

34,264,080 

60,440,750 

518,133 

"New  Jersey, 

16,076 

19,209 

60,742 

Pennsylvania, 

6,820,145 

8,464,882 

113,569 

Delaware, 

37,001 

802 

10,396 

Maryland, 

5,768,768 

4,910,746 

106,856 

Dist.  of  Columbia, 

753,923 

119,852 

17,711 

Virginia, 

4,778,220 

545,085 

47,536 

North  Carolina, 

387,484 

252,532 

31,682 

South  Carolina, 

10,036,769 

2,058,870 

23,469 

Georgia, 

6,862,959 

491,428 

16,536 

Alabama, 

12,854,694 

574,651 

14,577 

Louisiana, 

34,236,936 

10,673,190 

144,128 

Ohio, 

991,954 

4,915 

24,830 

Michigan, 

162,229 

148,610 

12,323 

Florida, 

1,858,850 

190,728 

7,288 

40  ADDITION*.  [SECT.  II. 

50.  The  appropriations  of  the  Government  of  the  United  States, 
for  1847,  were  as  follows :  for  the  Civil  and  Diplomatic  expenses 
4,442,790    dolls.;    for   the   Army   and   Volunteers    32,178,401 
dolls.;    for  the  Navy "9,307,958  dolls.;  for  the  Post  Office  De- 
partment 4,145,400  dolls.;  for  the  Indian  Department  1,364,204 
dolls.;    for  the  Military  Academy  124,906  dolls.;   for  building 
Steam  Ships  1,000,000  dolls. ;  for  Revolutionary  and  other  Pen- 
sions 1,358,700  dolls.;  for  concluding  Peace  with  Mexico  3,000, 

>00  dolls.;  for  Light  Houses  518,830  dolls.;  Miscellaneous 
540.243  dolls.  What  was  the  amount  of  all  the  appropriations? 

62.  It  may  sometimes  be  convenient  for  the  learner,  as  well 
as  gratifying  to  his  curiosity,  to  be  able  to  add  numbers  expressed 
by  the  Roman  Nutation. 

51.  A   man    paid    MDCCCLXXXIII    dollars    for    a    farm, 
DCCXXIIII   dollars    for   stock,   and  CCCLXVIIII   dollars  for 
tools  :  how  much  did  he  pay  for  all  ? 

Beginning  at  the  right  hand,  we  proceed  thus :  Operation. 

four  Is  and  four  Is  are  eight,  and  three  Is  make  MDCCCLXXXIII  dolls, 
eleven,  which  is  equal  to  two  Vs  and  I.  We  set  DCCXXIIII  dolls, 

down  the  I,  and  adding  the  two  Vs  to  one  V  CCCLXVIIII  dolls, 

makes  fifteen,  which  is  equal  to  X  and  V.  Set-  MMDCCCCLXXvT  dolls, 
ting  down  the  V,  we  count  in  the  X  with  the 

other  Xs,  and  find  they  make  seven  Xs  or  seventy,  which  is  expressed  by  L 
and  XX.  We  set  down  the  two  Xs,  and  adding  the  L  to  the  other  Ls,  il 
makes  three  Ls,  or  one  hundred  and  fifty,  which  is  expressed  by  C  and  L. 
Setting  down  the  L,  and  counting  the  C  with  the  other  Cs,  we  have  nine  Cs 
or  nine  hundred,  which  is  expressed  by  D  and  CCCC.  We  set  down  the  four 
Cs,  and  counting  the  D  with  the  other  Ds,  it  makes  three  Ds  or  fifteen  hun- 
dred, which  is  expressed  by  M  and  D.  We  set  down  the  D,  and  adding  the 
M  to  the  other  M,  we  have  two  Ms,  which  we  set  down  on  the  left  of  the  other 
letters.  Hence, 

63.  To  add  numbers  expressed  by  the  Roman  Notation. 

Beginning  at  the  right  hand,  count  all  the  letters  of  each  kind  to- 
gether ;  set  down  the  result,  and  carry  on  the  principle  that  five  Is 
make  one  V ;  two  Vs,  one  X ;  five  Xs,  one  L,  <fec.  . 

OBS.  The  teacher  can  extend  the  exercises  in  the  Roman  Notation  as  fa. 
as  he  may  deem  it  expedient.  A  single  example  is  sufficient  to  illustrate  the 
principle,  and  to  show  that  the  Roman  is  greatly  inferior  to  the  Arabic  method 
in  its  adaptation  to  business  calculations. 


ARTS.  62-66.]  SUBTRACTION.  41 

SECTION    III. 
SUBTRACTION. 

ART.  65.  Ex.  1.  A  merchant  bought  3*7  barrels  of  flour,  and 
afterwards  sold  12  of  them :  how  many  barrels  had  he  left? 
Solution. — 12  barrels  from  37  barrels  leave  25  barrels. 

Am.  25  barrels. 

OBS.  It  will  be  perceived,  that  the  object  in  this  example,  is  to  find  the  dif- 
ference between  two  numbers. 

66.  The -process  of  finding  the  difference  between  two  numbers 
is  called  SUBTRACTION. 

The  difference,  or  the  answer  to  the  question,  is  called  the 

Remainder. 

OBS.  1.  The  number  to  be  subtracted  is  sometimes  called  the  subtrahend^ 
and  the  number  from  which  it  is  subtracted,  the  minuend. 

2.  Subtraction,  it  will  be  perceived,  is  the  reverse  of  addition.     Addition 
unites  two  or  more  numbers  into  one  single  number ;  subtraction,  on  the  other 
hand,  separates  a  number  into  two  parts. 

3.  When  the  given  numbers  are  of  the  same  denomination ,  the  operation  is 
called  Simple  Subtraction.     (Art.  50.  Obs.) 

Ex.  2.  What  is  the  difference  between  5364  and  9387? 

Write  the  less  number  under  the  greater,          Operation. 
units  under  units,  tens  under  tens,  &c.     Then,  9387 

beginning  at  the  right  hand,  proceed  thus :  5364 

4  units  frtom  7  units  leave  3  units.      Write  4023    Rem. 

the  3  in  the  units'  place,  under  the  figure  subtracted.  6  tens 
from  8  tens  leave  2  tens ;  set  the  2  in  tens'  place.  3  hundred 
from  3  hundred  leave  0  hundred ;  we  therefore  write  a  cipher  in 
hundreds'  place.  5  thousand  from  9  thousand  leave  4  thousand; 
set  the  4  in  the  thousands'  place.  The  answer  is  4023. 

QUEST.— 66.  What  is  subtraction?  What  is  the  difference  or  answer  called  1  Ota 
What  is  the  number  to  be  subtracted  sometimes  called  ?  The  number  from  which  it  is 
subtracted  ?  Of  what  is  subtraction  the  reverse  ?  When  the  given  numbers  are  of  th« 
same  denomination,  what  is  the  operation  called  7 


42  SUBTRACTION.  [SECT.  III. 

67.  It  will  be  observed,  that  we  subtract  units  from  units, 
tens  from  tens,  &c. ;  that  is,  we  subtract  figures  of  the  same  order 
from  each  other.     This  is  done  for  the  same  reason  that  we  add 
figures  of  the  same  order  to  each  other.  (Art.  51.) 

OBS.  The  less  number  is  written  under  the  greater,  simply  for  convenience 
in  subtracting ;  and  units  are  placed  under  units,  tens  under  tens,  &c.,  to  avoid 
mistakes  which  might  occur  from  taking  different  orders  from  each  other. 

68.  It  often  happens  that  a  figure  in  the  lower  number  i 
Icurger  than  that   above  it,  and  consequently  cannot   be   taken 
from  it. 

Ex.  3.  What  is  the  difference  between  94  and  56  ? 

Analytic  solution.  It  is  manifest  that  we  cannot  take  6 
94=80  +  14  units  from  4  units,  for  6  is  larger  than  4. 
56  =  50+6  To  obviate  this  difficulty,  we  may  take 

Rem.  38=30+8  Iten  from  the  9  tens,  and  uniting  it 

with  the  4  units,  the  upper  number  will  become  8  tens  and  14 
units,  or  80  +  14.  Separating  the  lower  number  into  the  parts  of 
which  it  is  composed,  it  becomes  5  tens  and  6  units,  or  50+6. 
Now,  subtracting  as  in  the  last  example,  6  from  14  leaves  8,  50 
from  80  leaves  30.  The  answer  is  30  +  8,  or  38.  Or,  we  may 
simply  take  1  ten  from  the  9  tens,  and  adding  it,  mentally,  to  the 
4  units,  say  6  from  14  leaves  8  ;  set  the  8  under  the  figure  sub- 
tracted. Then,  having  taken  1  from  the  9  tens,  we  have  but  8 
left,  and  5  from  8  leaves  3.  The  answer  is  38. 

PROOF. — 38  +  56=94;  that  is,  the  sum  of  the  remainder  and 
smaller  number  being  equal  to  the  larger,  the  answer  is  right. 
Hence, 

69.  When  a  figure  in  the  lower  number  is  larger  than  that 
above  it ;  take  1  from  the  next  higher  order  in  the  upper  number, 
and  add  it  to  the  upper  figure  ;  from  the  sum  subtract  the  lower 
figure,  and  diminishing  the  next  upper  figure  by  1,  proceed  as 
before. 

OBS.  1.  The  process  of  taking  one  from  the  next  higher  order  and  adding  it 
to  the  figure  from  which  the  subtraction  is  to  be  made,  is  called  borrmoing  ten. 
It  is  the  reverse  of  carrying. 

QUEST. — 67.  What  orders  of  figures  do  you  subtract  from  each  other?  Why  no  ftA- 
tract  different  orders  from  each  other  ? 


ARTS.  67-71.]  SUBTR ACTION.  43 

2.  This  method  of  borrowu.g,  it  v^ill  be  seen,  does  not  ajfea  the  difference 
between  the  two  given  numbers ;  for,  it  is  simply  transposing  a  part  of  one 
order  to  another  order  in  the  same  number,  which,  it  is  obvious,  will  neither 
increase  nor  diminish  its  value. 

3.  It  may  be  asked,  how  can  we  take  one  from  the  figure  in  the  next  higher 
order,  when  that  figure  is  a  ciplier  ?    How  can  nothing  lend  anything,  and  how 
can  nothing  be  diminished  by  one  ?     The  explanation  of  this  apparent  contra- 
diction is  this :  when  the  next  figure  is  a  cipher,  we  go  to  the  next  higher 
column  still,  and  take  one,  which,  added  to  the  figure  in  the  next  lower  or.  ler, 
makes  ten ;  we  then  take  one  from  the  ten  and  add  it  to  the  upper  figure,  and 
proceed  as  before. 

7  O.  There  is  another  method  of  borrowing,  or  rather  of  pay- 
ing, which,  though  perhaps  less  philosophical  than  the  preceding, 
is  more  convenient  in  practice,  especially  when  the  figures  in  the 
next  higher  orders  are  ciphers.  Thus,  in  the  last  example,  adding 
10  to  the  upper  figure,  it  becomes  14,  and  6  from  14  leaves  8. 
Set  down  the  8  as  before.  JSFow,  instead  of  diminishing  the  next 
upper  figure  by  1,  if  we  add  1  to  the  next  figure  in  the  lower 
number  it  becomes  6  tens ;  and  6  from  9  leaves  3,  which  is  the 
same  as  5  from  8.  The  answer  is  38,  the  same  as  before.  Hence/ 
71.  When  a  figure  in  the  lower  number  is  larger  than  that 
above  it,  add  10  to  the  upper  figure,  and  to  compensate  this,  add 
1  to  the  .next  left  hand  figure  in  the  lower  number. 

OBS.  1.  This  method  of  borrowing  depends  on  the  self-evident  principle, 
that  if  any  two  numbers  are  equally  increased,  their  difference  will  not  be 
altered.  Thj^the  two  given  numbers  are  equally  increased  by  this  process, 
is  evident  i|Hkthe  fact  that  the  1  added  to  the  lower  number  is  of  the  next 
superior  order^o  the  10  added  to  the  upper  number,  and  is  therefore  equal 
to  it.  (Art.  35.) 

2.  The  reason  that  we  borrow  10,  instead  of  8,  or  12,  or  any  other  number, 
is  because  the  radix,  or  ratio  of  increase,  in  the  Arabic  notation,  is  10.     (Art. 
30.)     If  the  radix  of  the  system  were  8,  it  would  be  necessary  to  borrow  8; 
if  12.  it  would  be  necessary  to  borrow  12,  &c. 

3.  On  account  of  borrowing,  the  learner  will  perceive  it  is  always  necessary 
to  begin  to  subtract  at  the  right  hand. 

Ex.  4.  A  man  bought  a  house  for  23006  dollars,  and  sold  it  for 
21128  dollars  :  how  much  did  he  lose  by  his  bargain  ? 

Operation.  Proof. 

Cost     23006  dolls.  21128  Less  number. 

Rec'd.  21128  dolls.  1878  Remainder. 

Ans.     1878  dolls.  23006  Larger  number. 

T.H.  Q 


44  SUBTRACTION.  [SECT.    Ill, 

7  2.  From  the  preceding  illustrations  and  principles  we  derive 
the  following 

GENERAL  RULE  FOR  SUBTRACTION, 

I.  Write  the  less  number  under  the  greater,  so  tliat  units  may 
stand  under  units,  tens  under  tens,  <&c.  (Art.  67.  Obs.) 

]  I.  Beginning  at  the  right  hand,  subtract  each  figure  in  the  lower 
number  from  the  figure  above  it,  and  set  tJie  remainder  directly 
wider  the  figure  subtracted.  (Art.  71.  Obs.  3.) 

III.  When  a  figure  in  the  lower  number  is  larger  than  that  above 
it,  add  10  tt,  the  upper  figure  ;  then  subtract  as  before,  and  add  1 
to  tlie  next  figure  in  the  lower  number,  or  consider  the  next  upper 
figure  1  less  than  it  is.  (Arts.  69,  70.  Obs.  1,  2.) 

73.  PROOF. — Add  the  remainder  to  the  smaller  number  ;  and 
if  the  sum  is  equal  to  the  larger  number,  the  work  is  right. 

OBS.  This  method  of  proof  depends  upon  the  principle,  that  the  difference 
between  two  numbers  being  added  to  the  less,  the  sum  must  be  equal  to  the 
greater.  For,  the  difference  and  the  less  number  are  the  two  parts  into  which 
the  greater  is  separated,  and  the  whole  of  a  quantity  is  equal  to  the  sum  of  all 
its  parts.  (Ax.  11.) 

7  4.  Second  Method. — Subtract  the  remainder  from  the  greater 
of  the  two  given  numbers  ;  and  if  the  difference  is  equal  to  the 
less  number,  the  work  is  right. 

75.  Third  Method. — Cast  the  9s  out  of  the  larger  number,  and  place  the 
excess  at  the  right.  Next,  cast  the  9s  out  of  the  smaller  number,  and  also 
out  of  the  remainder;  then  cast  the  9s  out  of  the  sum  of  these  two  excesses, 
and  if  this  last  excess  is  the  same  as  the  excess  of  the  larger  number,  the  work 
may  be  supposed  to  be  right.  Thus, 

Ex.  5.  From  7843  Excess  of  9s  in  the  greater  number  is  4 
Take  5675       "         «         «     less  "      is  5  > 

Rcm.  21G8       "         "        "      remainder  is         8  \  Now,  8-}-5=l3, 
and  the  excess  of  9s  in  13  is  4,  the  same  as  that  of  the  greater  number. 

QUEST. — 72.  How  do  you  write  numbers  for  subtraction  ?  Why  write  the  less  number 
under  the  greater  ?  Why  place  units  under  units,  &c.  1  Where  do  you  begin  to  subtract  ? 
When  a  figure  in  the  lower  line  is  larger  than  that  above  it,  how  do  you  proceed  1  What 
is  meant  by  borrowing  ten  ?  How  many  methods  of  borrowing  are  mentioned  ?  Illustrate 
the  first  method  upon  the  black-board.  How  does  it  appear  that  this  method  of  borrowing 
does  not  attest  the  difference  between  the  two  given  numbers?  Explain  the  second  me- 
thod. Upon  what  principle  does  this  method  depend  1  Why  do  yoi  borrow  10,  instead 
of  H,  or  12,  or  any  other  number?  Why  do  you  nei;in  to  subtract  at  the  right  hand  1 
73.  How  is  subtraction  proved  ?  O6.<r.  Upon  what  principle  does  thi?  method  of  proof  de- 
pend 1  Can  subtraction  be  proved  by  any  other  methods  ? 


ARTS.  72—76.]  SUBTRACTION.  45 

Note. — This  method  of  proof  depends  on  the  same  property  of  the  number 
9,  as  that  in  addition.  (Art.  58.  Note.)  For,  since  the  sura  of  the  smaller 
number  and  remainder  is  equal  to  the  larger  number,  it  follows  that  the 
excess  of  9s  in  the  larger  number  must  be  equal  to  the  excess  of  9s  in  the 
remainder  and  smaller  number  together. 

EXAMPLES    FOB    PRACTICE. 

76.  Ex.  1.  A  merchant  bought  a  ship  for  35270  dollars,  and 
r  >ld  it  for  42365  dollars  :  how  much  did  he  make  by  his  bargain  ? 

2.  A  miller  bought  46235  bushels  of  wheat,  and  ground  17251 
bushels  of  it :  how  many  bushels  had  he  left  ? 

3.  A  speculator  laid  out  50000  dollars  in  wild  land,  and  after- 
wards sold  it  at  a  loss  of  19046  dollars  :  how  much  did  he  get  for 
his  land  ? 

4.  A  man  owning  a  block  of  buildings  worth  155265  dollars, 
keeps  it  insured  for  109240  dollars  :  how  much  would  he  lose  in 
case  the  buildings  should  be  destroyed  by  fire  ? 

5.  The  distance  from  the  Earth  to  the  Sun  is  95000000  of 
miles  ;  the  distance  of  Mercury  is  only  37000000 :  how  far  is 
Mercury  from  the  Earth  ? 

6.  The  imports  of  Massachusetts  in   1840,   were   16,513,858 
dollars,  the  exports  were  10,186,261  dollars:  what  was  the  ex- 
cess of  her  imports  over  her  exports  ? 

7.  The  imports  of  New  York  in  1840,  were  60,440,750  dol- 
lars, the  exports  were  34,264,080  dollars :  what  was  the  excess 
of  her  imports  over  her  exports  ? 

8.  The  imports  of  Pennsylvania  in  1840,  were  8,464,882  dol- 
lars, the  exports  were  6,820,145  dollars  :  what  was  the  excess  of 
her  imports  over  her  exports  ? 

9.  The  imports  of  South   Carolina  in   1840,   were   2,058,870 
dollars,  the  exports  were  10,036,769  dollars :  what  was  tLe  ex- 
cess of  her  exports  over  her  imports  ? 

10.  The  imports  of  Alabama  in  1840,  were  574,651  dollars, 
the  exports  were  12,854,694  dollars  :  what  was  the  excess  of  her 
exports  over  her  imports  ? 

11.  The  imports  of  Louisiana  in  1840,  were  10,673,190  dol- 
lars, the  exports  were  34,236,936  dollars :  what  was  the  excess' 
of  her  exports  over  her  imports  ? 


46  SUBTRACTION.  [SECT.  III. 

12.  The  tonnage  of  the  United  States  in  1842,  was  2069857, 
w  1846  it  was  2500000  :  what  was  the  increase  in  4  years? 

13.  14.                            15. 

From  253760  3856031               54903670 

Take   104523  462702                   504089 

16.  9876102—1050671.  28.   10000000—999999. 

17.  4006723—5001.  29.  99999999—100000. 

18.  3601900—1000000.  30.   83567000—438567, 

19.  5317004 — 3565.  31.  40600056 — 7632. 

20.  1000000 — 456321.  32.  56409250 — 1057245. 

21.  2035024—27040.  33.  20030000 — 72534. 

22.  45563075—460001.  34.  83175621—5256360. 

23.  67030001—300452.  35.  70301604—250041. 

24.  73256300—436020.  36.  60050376—6849005. 

25.  56037431 — 735671.  37.  34200591 — 8888888. 

26.  80200430—250.  38.  87035762 — 753017. 

27.  96531768—873625.  39.  95246300 — 9438675. 

40.  From  6764+3764  take  6500+2430. 

41.  From  2890+8407  take  4251+3042. 

42.  From  7395+4036  take  8297+1750. 

43.  From  8404+7296  take  3201 — 1562. 

44.  From  6008+9270  take  5136—2352. 

45.  From  9234  +  6850  take  9320—4783. 

46.  From  8564—2573  take  4431—1735. 

47.  From  7284—5362  take  6045—5729. 

48.  From  9561—4680  take  7352—6178. 

49.  From  8630 — 1763  take  2460  +  1743. 

50.  From  7561—2846  take  1734  +  2056. 

51.  From  9687—3401  take  3021  +  1754. 

52.  A  man  having  55000  dollars,  paid  7520  dollars  for  a  house,, 
8260   dollars  for  furniture,   2375  dollars  for  a  library,  and  in- 
rested  the  balance  in  bank  stock :  how  much  stock  did  he  buy  ? 

53.  A  gentleman  worth  163250  dollars,  bequeathed  15200  dol- 
lars apiece  to  his  two  sons,  16500  dollars  to  his  daughter,  and  to 
his  wife  as  much  as  to  his  three  children,  and  the  remainder  to  a 
hospital ;  how  much  did  his  wife  receive,  and  how  much  the  hos« 
pital  ? 


ARTS.  76,  77.]  SUBTRACTION.  47 

54.  A  man  bought  three  farms  :  for  the  first  he  paid  5260  dol- 
lars, for  the  second  3585,  and  for  the  third  us  much  as  for  the 
first  two.     He  afterwards  sold  them  all  for  15280  dollars:  did  he 
make  or  lose  by  the  operation ;  and  how  much  ? 

55.  What  number  is  that,  to  which  3425  being  added,  the  sum 
will  be  175250? 

56.  A  man  being  asked  how  much  he  was  worth,  replied,  if 
you  will  give  me  325263  dollars,  I  shall  have  two  millions  of  dol- 
lars :  how  much  was  he  worth  ? 

57.  A  jockey  gave  150  dollars  for  a  horse,  and  meeting  an  ac- 
quaintance swapped  with  him,  giving  37  dollars  to  boot ;  meeting 
another,  he  swapped  and  received  28  dollars  to  boot ;  he  finally 
swapped  again  and  gave  78  dollars  to  boot,  and  then  sold  his  last 
horse  for  140  dollars :  how  much  did  he  lose  by  all  his  bargains  ? 

58.  A  speculator  gained  3560  dollars,  and  afterwards  lost  2500 
dollars;   at  another  time  he  gained  6283  dollars,  and  then  lost 
3450  dollars :  how  much  more  did  he  gain  than  lose  ? 

59.  A  man  bought  a  house  for  MDCCCCXXXVII  dollars,  and 
sold  it  for  DCXVIIII  dollars  less  than  he  gave:  how  much  did 
he  sell  it  for  ? 

We  perceive  that  the  IIII  in  the  lower  number  Operation. 

cannot  be  taken  from  II  in  the  upper  number;  MDCCCCXXXVII  dolls, 
we  therefore  borrow  a  V,  which  added  to  the  II,  DCXVIIII  dolls, 

makes  IIIIIII ;  then  IIII  from  IIIIIII,  leaves  Ans.  MCCCXVIII  dolls. 
Ill,  which  we  set  down.  Now  since  we  have 

borrowed  the  V  in  the  upper  number,  there  are  no  Vs  left  from  which  we  can 
take  the  V  in  the  lower  number.  We  must  therefore  borrow  an  X ;  but  X  ia 
equal  to  VV;  and  V  from  VV  leaves  V,  which  we  set  down.  Having  bor- 
rowed an  X  from  the  upper  number,  there  are  but  XX  left,  and  X  from  XX 
leaves  X.  C  from  CCCC  leaves  CCC.  D  from  D  leaves  nothing.  And 
nothing  from  M  leaves  M.  Hence, 

77.  To  subtract  numbers  expressed  by  the  Roman  Notation. 

Write  the  less  number  under  Hie  greater ;  then,  beginning  at  the  right  hand, 
take  the  number  in  the  lower  line  from  that  expressed  by  the  same  tetters  in  the 
upper  line,  and  set  the  remainder  below.  If  the  number  in  the  lower  line  it 
larger  than  that  expressed  by  the  same  letters  in  the  upper  line,  bwrmt)  a  letter 
next  hig/tcr  and  add  it  to  the  number  in  the  upper  line ;  then  subtract  as  bcfcre, 
observing  to  pay  when  you  borrow  as  in  subtraction  of  figures.  (Art.  72.) 

OBS.  Other  examples  expressed  by  the  Roman  Notation,  can  be  added  by 
the  teacher,  if  deemed  expedient. 


48  MULTIPLICATION  |  SECT.   IV 


SECTION    IV. 

MULTIPLICATION. 

ARTT.  79*  Ex.  1.  What  will  3  melons  cost,  at  15  cents  apiece 

Analysis. — If  1  melon  costs  15  cents,  3  melons  will  cost  3  times 
15  cents;  and  3  times  15  cents  are  45  cents.  Ans.  45  cents. 

2.  What  will  4  sleighs  cost,  at  21  dollars  apiece  ? 

Analysis. — Reasoning  as  before,  if  1  sleigh  costs  21  dollars,  4 
sleighs  will  cost  4  times  as  much;  and  4  times  21  dollars  are 
84  dollars.  Ans.  84  dollars. 

OBS.  It  is  obvious  that  3  times  15  cents  is  the  same  as  15  cents-f-15  cents 
-{-15  cents,  or  15  cents  added  to  itself  3  times ;  and  4  times  21  dollars  is  the 
same  as  21  dolls.-f-21  dolls.-j-21  dolls.-f  21  dolls.,  or  21  dollars  added  to  itself 
4  times. 

80.  Thin  repeated  addition  of  a  number  or  quantity  to  itself,  is 
called  MULTIPLICATION. 

The  number  to  be  repeated,  or  multiplied,  is  called  the  Multi- 
plicand. 

The  number  by  which  we  multiply,  is  called  the  multiplier; 
and  shows  Iww  many  times  the  multiplicand  is  to  be  repeated. 

The  number  produced,  or  the  answer  to  the  question,  is  called 
the  product.  Thus,  when  we  say,  8  times  12  are  96,  8  is  the 
multiplier,  12  the  multiplicand,  and  96  the  product. 

81.  The  multiplier  and  multiplicand  together  are  often  called 
factors,  because  they  make  or  produce  the  product. 

OBS.  1.  The  term  factor  is  derived  from  a  Latin  word  which  signifies  a 
tgcnt,  a  doer,  or  prodiiccr. 

2.  When  the  multiplicand  denotes  things  of  one  denomination  only,  the  ope- 
ration is  called  Simple  Multiplication. 

GUEST.— 80.  What  is  multiplication  7  What  is  the  number  to  be  repeated  called? 
What  the  number  by  which  we  multiply?  What  does  the  multiplier  show!  What  la 
the  number  produced  called?  81.  What  are  the  multiplicand  and  multiplier  togetbei 
vailed  1  Why  1  Obs.  What  does  the  term  factor  signify  1 


.  79~82.| 


MULTIPLICATION. 


49 


MULTIPLICATION    TABLE. 


1 

2 

3 

4 

5 

6|     7 

S 

9i   10 

11     12 

13 

14 

15 

lo    17 

18 
36 

19 

_20 
40 

_|_4 

6 

8 

10 

12 

14 

16 

18    20 

22 

24 

26 

28 

30 

32 

31 

38 

9 

12 

15 

18 

21 

24 

27 

3J 

33 

3:1 

39 

42 

45 

4* 

51 

54 

5; 

60 

4 

12 

16 

21 

24 

28|   32 

36 

40 

44 

48 

52 

56 

60 

64 

68 

72 

76 

80 

loo 

5 

U 

15   2~J 

25 

30 

35 

1  . 

45 

50 

55 

6U 

i  i."; 

70 

75 

so 

85 

9d 

95 

r> 

12 

18  1  24 

30 

3J 

42  1  48 

54 

60 

66 

72 

78 

HI 

90 

96 

102 

108 

II4J120 

7 
8 

11 
16 

211  28 

35 

42 

49 

56 

63 

70 

77 

HI 

91 

98 

105 

Ii2  119 

126 

133 

140 

24 

32 

40 

48 

56 

64 

72 

H  ! 

88 

96 

104 

112 

120  128  136 

144 

1  52 

160 

9|  18 

27 

:!  ; 

45 

54 

63 

'  - 

81 

g 

9d 

108 

117 

12o  135  I44JI53 

162;17l|l8- 

10 

20 

30 

1  1 

5!) 

60 

70 

BO 

90 

1  .0 

110 

120il3;) 

140 

150  1160  '170 

180  UK)  20L 

11 

72 

*>-> 

**  1 

33 

44 

55 

I'M) 

77    88 

99 

111! 

121  132  143 

154  MS  17(i  J87|l!H 

20!  I 

220 

36 

48 

60 

72 

84 

96 

108!  12) 

1321144  156 

16S 

180  1112  204 

2ifii2-J8 

240 

13 

26 

39 

52 

6.1 

78 

yi 

I04|117|13.j 

143  15  ij  109  182 

l:'5:2o8'221 

234^247 

Si 

14 

28|  42 

56 

7,1 

84 

U8|ll2!l26|!4!) 

1  54  1  1  1  18  :  1  82  i  I'.i:  i  1  2  1  0  '  224  \  2:<S 

252  26(i 

15 
16 

31)1  45 

60'   75 

5)0|  105 

120;  135'  150  1-.5  180  195  210  225  2401255 

2  70  !  285 

300 

32  48 

64  1  80 

9> 

112 

I2SJ144  I6o|l7<>jl!)2  2(Hi224i24  »  25!>!2;2 

288;  304 

320 

17 

34 

51 

66    85 

102 

119 

136153  170 

1  87  1  -204  22  1  238  ;  255  272  '  2-<!» 

:M)t),:i23 

340 
3~iiO 

-181  36 

51 

72 

90 

108 

l~2ii 

144!  1(52  IriO!l'.)8  21(5  234  252  2iO  288,306 

324  342 

19  38 

57 

7;;i  95 

114 

L33 

152  I'll  190  '209  228  247;  2'  16  285  1  304 

323  342  3til 

380 

2~jl  40i  60  1  80ilOJ  120 

140 

160il80  200  1  2  20,  240  260  280  1  300  320 

340,3GO;380J4UO 

Note. — This  Table  was  invented  by  Pythagar-as,  and  is  therefore  sometimes 
called  the  Pythagorean  Table. 

The  pupil  will  find  assistance  in  learning  the  Multiplication  Table  ty  ob 
serving  the  following  particulars. 

1.  The  several  results  of  multiplying  by  10  are  formed  by  simply  adding  a 
cipher  to  the  figure  that  is  to  be  multiplied.     Thus,  10  times  2  are  20,  10  times 

3  are  30,  &c. 

2.  The  results  of  multiplying  by  5  terminate  in  5  and  0,  alternately.     Thus, 
5  times  1  are  5,  5  times  2  are  10,  5  times  3  are  15,  &c. 

3.  The  first  nine  results  of  multiplying  by  1 1  are  formed  by  repeating  the 
figure  to  be  multiplied.     Thus,  11  times  2  are  22 ;  11  times  3  are  33,  &c. 

4.  In  the  successive  results  of  multiplying  by  9,  the  right  hand  figure  regu- 
larly decreases  by  1,  and  the  left  hand  figure  regularly  increases  by  1.     Thus, 
9  times  2  are  18;  9  times  3  are  27;  9  times  4  are  36,  &c. 

82.  Multiplying  by  1,  is  taking  the  multiplicand  once:  thus, 

4  multiplied  by  1  =  4. 

Multiplying  by  2,  is  taking  the  multiplicand  twice :  thus,  2  times 
4,  or  4+4  =  8. 

Multiplying  by  3,  is  taking  the  multiplicand  three  times :  thus, 
3  times  4,  or  4+4 -f- 4 =12,  &c.  Hence, 


QUEST.—  82,  What  is  it  to  multiply  by  1  ?    By  2  1 
5 


By  3  1 


50  MULTIPLICATION.  [SfiCT.  IV 

Multiplying  ly  any  whole  number,  is  talcing  the  multiplicand  as 
many  times,  as  there  are  units  in  the  multiplier. 

The  application  of  this  principle  to  fractional  multipliers  will 
be  illustrated  under  fractions. 

OBS.  1.  From  the  definition  of  multiplication,  it  is  manifest  that  the  product 
is  of  the  same  kind  or  denomination  as  the  multiplicand :  for,  repeating  c  num- 
ber or  quantity  does  not  alter  its  nature.  Thus,  if  we  repeat  dollars,  they  ara 
still  dollars ;  if  we  repeat  yards,  they  are  still  yards,  &c.  Consequently,  if  the 
multiplicand  is  an  abstract  number,  the  product  will  be  an  abstract  number;  if 
money,  the  product  will  be  money ;  if  barrels,  barrels,  &c. 

&  Every  multiplier  is  to  be  considered  an  abstract  number.  In  familiar 
*«nguage  it  is  sometimes  said,  that  the  price  multiplied  by  the  weight  will  give 
the  value  of  an  article ;  and  it  is  often  asked  how  much  25  cents  multiplied  by 
i35  cents,  &c.,  will  produce.  But  these  are  abbreviated  expressions,- and  are 
liable  to  convey  an  erroneous  idea,  or  rather  no  idea  at  all.  If  taken  literally, 
they  are  absurd ;  for  multiplication  is  repeating  a  number  or  quantity  a  certain 
number  of  times.  Now  to  say  that  the  price  is  repeated  as  many  times  as  the 
given  quantity  is  lieavy,  or  that  25  cents  are  repeated  25  cents  times,  is  non- 
sense. But  we  can  multiply  the  price  of  1  pound  by  a  number  equal  to  the 
number  of  pounds  in  the  weight  of  the  given  article,  and  the  product  will  be 
the  value  of  the  article.  We  can  also  multiply  25  cents  by  the  number  25 ; 
that  is,  repeat  25  cents  25  times,  and  the  product  is  625  cents.  Construed  in 
tljis  manner,  the  multiplier  becomes  an  abstract  number,  and  the  expressions 
have  a  consistent  meaning. 

Ex.  3.  What  will  6  houses  cost,  at  2341  dollars  apiece  ? 

Write  the  numbers  on  the  slate  as  Operation. 

in  the  margin,  and  beginning  at  the  2341  Multiplicand, 

right  hand,  proceed  thus :  6  times  1  6  Multiplier, 

unit  are  6  units ;  write  the  6  under  the  Ans.  14046  Dollars, 
figure  multiplied.  6  times  4  tens  are  24  tens  ;  set  the  4  or  right 
hand  figure  under  the  figure  multiplied,  and  cany  the  2  or  left 
hand  figure  "o  the  next  product  figure,  as  in  addition.  (Art.  52.) 
6  times  3  hundreds,  are  18  hundreds,  and  2  to  carry  make  20  hun- 
dreds ;  set  the  0  under  the  figure  multiplied,  and  carry  the  2  to 
the  next  product  as  before.  6  times  2  thousands  are  12  thou- 
sands, and  2  to  carry  make  14  thousands.  Since  there  are  no 

QITKST. — What  is  it  to  multiply  by  any  whole  number  7  Obs.  Of  what  denomination  is 
the  prodact?  How  does  this  appear  7  What  must  every  multiplier  be  considered  7  Caj 
you  multiply  by  a  given  weight,  a  measure,  or  a  sum  of  money  ? 


ART.  83.]  MULTIPLICATION.  51 

more  figures  to  be  multiplied,  set  down  the  14  in  full  as  in  addi- 
tion. (Art.  53.  Obs.  1.)     The  product  is  14046  dollars. 

83*  The  product  of  any  two  numbers  will  be  the  same,  which- 
ever factor  is  taken  for  the  multiplier.  Thus, 

If  an  orchard  contains  5  rows  of  trees,  and  ******* 
each  row  has  7  trees,  as  represented  by  the  ******* 
stars  in  the  margin,  it  is  evident  the  whole  ******* 
umber  of  trees  is  equal  either  to  the  number  ******* 
of  stars  in  a  horizontal  row  repeated  Jive  times,  ******* 
or  to  the  number  of  stars  in  a  perpendicular  row  repeated  seven 
times,  viz.  35.  For,  7X5  =  35,  also  5X7  =  35. 

OBS.  1.  It  is  more  convenient  and  therefore  customary  to  place  the  larger  num 
her  for  the  multiplicand,  and  the  smaller  for  the  multiplier.  Thus,  it  is  easier 
to  multiply  8468946  by  3,  than  it  is  to  multiply  3  by  8468946,  but  the  product 
would  be  the  same.  * 

i* 

Ex.  4.  What  will  237  coaches  cost,  at  675  dollars  apiece? 

Since  it  is  not  convenient  to  multi-  Operation. 

ply  by  237  at  once,  we  multiply  first  675  Multiplicand, 

by  the  7  units,  next  by  the  3  tens,  237  Multiplier, 

then  by  the  2  hundreds,  and  place  4725  cost  7   coaches, 

each  result  in  a  separate  line,  with  2025*  cost  30      " 

the  first  figure  of  each  line  directly  1350**  cost   200    " 

under  that  by  which  we  multiply.  159975  cost  237    " 
Finally,  adding  these  results  togeth- 
er, units  to  units,  &c.,  we  have  159975  dollars,  which  is  the  whole 
product  required.     (Ax.  1 1 .) 

Note. — When  the  multiplier  contains  more  than  one  figure,  the  several  pro- 
ducts of  the  multiplicand  into  tne  Keparate  figures  of  the  multiplier,  are  called 
partial  products. 

OBS.  2.  The  reason  for  placing  the  first  figure  of  the  several  partial  products 
undsr  the  figure  by  which  we  multiply,  is  to  bring  the  same  orders  under  each 
other,  and  thus  prevent  mistakes  in  adding  them  together.  (Art.  51.) 

3.  The  several  partial  products  are  added  together  for  the  obvious  purpose 
of  finding  the  whole  product  or  answer  required.  (Ax.  11.) 

QUEST.— 83.  Does  it  make  any  difference  with  the  result,  which  of  the  given 
i»  taKen  for  the  multiplier  1     Obs,  Which  is  usually  taken  ?    Why  1 
3* 


52  MULTIPLICATION.  .          [SECT.  IV. 

84*  The  principle  of  carrying  the  tens  in  multiplication  is  the 
same  as  in  addition,  and  may  be  illustrated  in  a  similar  manner. 
(Art.  53.)  Thus, 

Ex.  5.     9382  Mult'd.  Or,  separating  the  multiplicand  into 

7  Mult'r.          the  orders  of  which  it  is  composed, 
14= units,  9382  =  9000  +  300  +  80  +  2, 

56*=tens,  and  9000X7  =  63000 

21**=hunds.  300X7=   2100 

63***=thou.  80X7=     560 

65674  Product.  2x7  = 14 

Adding  these  results  together,  we  have       65674  Ans. 
OBS.  The  reason  for  always  beginning  to  multiply  at  the  right  hand  of  the 
multiplicand,  is  that  we  may  carry  the  tens  as  we  proceed  in  the  operation. 

85.  From  this  illustration  it  will  be  observed  that  units  mul- 
tiplied into  units  produce  units  ;  tens  into  unjts,  or  units  into  tens, 
produce"  tens  ;  (Art.  83  f)  hundreds  into  units,  or  units  into  hun- 
dreds, produce  hundreds,  &c.  Hence, 

86»  When  units  are  multiplied  into  any  order  whatever,  tlie 
product  will  always  be  of  the  same  order  as  the  other  figure. 

And  universally,  the  product  of  any  two  integers  is  of  the  order 
next  less  than  that  denoted  by  the  sum  of  the  orders  of  tlie  two  given 
figures.  Thus,  hundreds  into  tens  produce  thousands,  or  the  4th 
^rder,  which  is  one  less  than  the  sum  of  the  two  given  orders. 

OBS.  When  the  multiplier  contains  more  than  one  figure,  it  is  customary  to 
begin  to  multiply  with  its  units'  figure.  The  result  however  will  be  the  same, 
if  we  begin  with  its  hundreds  or  any  other  order  of  the  multiplier,  and  place 
the  first  figure  of  the  partial  products,  so  that  the  same  orders  shall  stand 
under  each  other. 


First  Operation. 
1357 

Second  Operation. 
1357 

3574 

3574 

407T~ 

4071 

6785 

6785 

9499 

9499 

5428 

5428 

4849918.  Prod. 

4849918.  Prod. 

QUEST.— 85.  What  do  units  into  units  produce  ?    Units  into  tens,  or  tens  into  units  1 


ARTS.  84-89.]  MULTIPLICATION.  53 

Ex.  6.  What  is  the  product  of  5690  into  3008  ? 

After  multiplying  by  the   8   units,  we  next          Operation. 
multiply  by  the  8  thousands,  since  there  arc  no  5G90 

tens  nor  hundreds  in  the  multiplier,  and  place  3008 

the  first  figure  of  this  partial  product  under  the  45520 

figure  3  by  which  we  are  multiplying.  17070 

17115520  Ans. 

87.  From  the  preceding  illustrations  and  principles  we  de- 
rive the  following 

GENERAL   RULE   FOR  MULTIPLICATION. 

I.  When  the  multiplier  contains  but  one  figure. 

Write  the  multiplier  under  the  multiplicand,  units  under  units, 
tens  under  tens,  d'c.  (Art.  83.  Obs.  1.) 

Beyin  at  the  right  hand  and  multiply  each  figure  of  the  mul- 
tiplicand by  the  multiplier,  setting  doivn-  the  result  and  carrying  as 
in  addition.  (Art.  84.  Obs.) 

II.  When  the  multiplier  contains  more  than  one  figure. 
Multiply  each  figure  of  the  multiplicand  l)y  each  fiyure  of  tJie 

multiplier  separately,  beyinniny  with  the  units,  and  write  the  par- 
tial products  in  separate  lines,  placing  the  first  figure  of  each  line  di~ 
rectly  under  the  fiyure  by  which  you,  multiply.   (Art.  8G.  Obs.  2.) 
Finally,  add  the  several  partial  products  toyetJter,  and  tJie  sum 
will  be  the  whole  product.  (Art.  83.  Obs.  3.) 

88.  PROOF. — Multiply    the   multiplier   by  the  multiplicand, 
and  if  the  product  thus  obtained  is  tlie  same  as  the  other  product, 
the  work  is  supposed  to  be  right. 

OBS.  This  method  of  proof  depends  upon  the  principle,  that  the  product  of 
any  two  numbers  is  the  same,  whichever  is  taken  for  the  multiplier.  (Art.  83.) 

89.  Second  Method. — Add  the  multiplicand  to  itself  as  many 

QUEST.— 86.  When  units  are  multiplied  into  any  order,  what  order  is  the  product? 
When  any  two  integers  are  multiplied  together,  of  what  order  is  the  product  ?  87.  How 
do  you  write  the  numbers  for  multiplication  ?  When  the  multiplier  contains  liut  one  fig 
Ore,  how  proceed  ?  Why  begin  at  the  right  hand  of  the  multiplicand  1  When  the  multi 
pUer  contains  more  than  one  figure,  how  proceed?  What  is  meant  l>y  partial  products  1 
Why  place  the  first  figure  of  each  partial  product  under  the  figure  by  which  yon  multi- 
ply? What  is  to  be  done  with  the  partial  products?  Why  add  the  several  partial  pro 
dacts  together?  Why  should  this  give  the  whole  product?  88.  How  is  multipUcaitoa 
proved  ?  Obs.  On  what  principle  does  this  proof  depend  ? 

5* 


54  MULTIPLICATION.  [SECT.  IV. 

times  as  there  are  units  in  the  multiplier,  and  if  the  product  ob- 
tained is  equal  to  the  amount,  the  work  is  right. 

jyole^ — when  the  multiplier  is  small,  this  is  a  very  convenient  mode  of  proof. 

90.  Third  Method. — Cast  the  9s  out  of  the  multiplicand  and 
multiplier ;  multiply  their  remainders  together,  and  casting  the 
9s  out  of  their  product,   set  down  the  excess  ;   then   cast  the 
9s  out  of  the  answer  obtained,  and  if  this  excess  be  the  same  as 
that  obtained  from  the  multiplier  and  multiplicand,  the  work  may 
be  considered  right. 

Ex   7.  Multiply  565  by  356. 

Operation.  Proof. 

565     The  excess  of  9s  in  the  multiplicand  is  7. 
356  "          9s     "       multiplier  is      5. 

3390    7X5  =  35;  and  the  excess  of  9s  is      8. 
2825 
1695 
Prod.  201140.     The  excess  of  9s  in  the  Ans.  is  also  8. 

91.  Fourth  Method. — Divide  the  product  by  one  of  the  fac- 
tors, and  if  the  quotient  thus  arising  is  equal  to  the  other  factor, 
the  work  is  right. 

Note. — This  method  of  proof  supposes  the  learner  to  be  acquainted  with 
division  before  he  commences  this  work.  (Art.  57.  Note.)  It  is  simply  re- 
versing the  operation,  and  must  obviously  lead  us  back  to  the  number  with 
which  we  started :  for,  if  a  number  is  both  multiplied  and  divided  by  the  same 
number,  its  value  will  not  be  altered.  (Ax.  9.) 

92.  Fifth  Method.*— First,  cast  the  11s  out  of  the  multiplicand  and  multi- 
plier; multiply  their  remainders  together,  cast  the  11s  out  of  the  product,  and 
set  down  the  excess ;  then  cast  the  11s  out  of  the  answer  obtained,  and  if  the 
excess  is  the  same  as  that  obtained  from  the  multiplier  and  multiplicand,  the 
work  is  right. 

Note. — 1.  This  method  depends  on  a  peculiar  property  of  the  number  11. 
For  its  further  development  and  illustration,  see  Art.  161.  Prop.  18. 

2.  To  cast  the  11s  out  of  a  number,  begin  at  the  right  hand,  mark  the  alter- 
nate figures;  then  from  the  sum  of  the  figures  marked,  increased  by  11  if 
necessary,  take  the  sum  of  those  not  marked,  and  the  remainder  will  be  the 
excess  required.  Thus  to  cast  the  11s  out  of  39475025,  mark  the  alter- 
figuresj  beginning  at  the  right  hand,  39475025,  then  the  sum  of 

QUEST. — Can  multiplication  be  proved  by  any  other  methods? 
*  Leslie  s  Philosophy  of  Arithmetic. 


AKTS.  90-93.]  MULTIPLICATION.  55 

5_l_0-{-7-|-9=21.     Again,  the  sum  of  the  others,  viz:  2+5-|-4-}-3= 14.    Now, 
£1—14=7,  the  excess  of  11s. 

Or,  as  soon  as  the  sum  is  11  or  over,  we  may  drop  the  11,  and  add  the  re- 
mainder to  the  next  digit.     Thus,  5  and  7  are  12;  dropping  the  11,  1  and  9 
are  10.     Again,  2  and  5  are  7,  and  4>.are  11 ;  drop  the  11,  and  there  are  3  left. 
Now,  10 — 3=7,  the  same  excess  as  before. 
Ex.  8.  Multiply  237956  by  3728. 

Operation.  Proof. 

237956     Excess  of  lls  is  4.  >      Now,  4X10=40;  the  excess  of  1U 

3728        "  "        10.  $         in  40  is  7. 

Ans.  887099908    Excess  of  lls  in  the  answer  is  also  7. 

EXAMPLES    FOR    PRACTICE. 

93*  Ex.  1.  What  will  435  acres  of  land  cost,  at  57   dollars 
per  acre  ? 

2.  What  cost  573  oxen,  at  63  dollars  per  head  ? 

3.  What  cost  1260  tons  of  iron,  at  45  dollars  per  ton  ? 

4.  If  a  man  can  travel  248  miles  in  a  day,  how  far  can  he 
travel  in  365  days  ? 

5.  If  an  army  consume  645  pounds  of  meat  in  a  day,  how 
much  will  they  consume  in  1 1 5  days  ? 

6.  If  1250  men  can  build  a  fort  in  298  days,  how  long  would 
it  take  1  man  to  do  it  ? 

7.  How  many  rods  is  it  across  the  Atlantic  Ocean,  'allowing 
320  rods  to  a  mile,  and  the  distance  to  be  3000  miles  ? 

8.  What  is  the  product  of  463X45? 

9.  What  is  the  product  of  348X62? 

10.  What  is  the  product  of  793X86  ? 

11.  What  is  the  product  of  75X42X56  ? 

12.  What  is  the  product  of  7198X216  ? 

13.  31416X175.  22.  8320900X1328. 

14.  8862X189.  23,   17500X732. 

15.  7071X556.  24.  15607X3094. 

16.  93186X4455.  25.  742215^X468. 

17.  40930X779.  26.  9264397x9584. 

18.  12345X686.  27.  46873J9X1987. 

19.  46481X936.  28.  95073*0X7071. 

20.  16734X708.  29.  39948123X6007. 

21.  7^5X7575.  30.  73885246X6079. 


56  MULTIPLICATION.  [SECT.   IV. 

31.  51902468X5008.  37.  58763718X6754. 

32.  57902468X5080.  38.   73084163X7584. 

33.  57902468X5800.  39.  144X144X144. 

34.  12481632X1509.  40.   3851X3851X3851. 

35.  79068025X1386.  41.  79094451X764094. 

36.  92948789X7043.  42.  89548050X972800. 

CONTRACTIONS  IN  MULTIPLICATION. 

94.  The  general  rule  is  adequate  to  the  solution  of  all  exam 
pies  that  occur  in  multiplication.     In  many  instances,  however, 
by  the  exercise  of  judgment  in  applying  the  preceding  principles, 
the  operation  may  be  very  much  abridged. 

95.  Any  number  which  may  be  produced  by  multiplying  two 
or  more  numbers  together,  is  called  a  Composite  Number. 

Thus,  4,  15,  21,  are  composite  numbers;  for  4=2X2;  15  = 
5X3;  21  =  7X3. 

OBS.  1.  The  factors  which,  being  multiplied  together,  produce  a  composite 
number,  are  sometimes  called  the  component  parts  of  the  number. 

y.  The  process  of  finding  the  factors  of  which  a  given  number  is  composed, 
is  called  resolving  tJie  number  into  factors. 

Ex.  1.  Resolve  9,  10,  14,  22,  into  their  factors. 

2.  What  are  the  factors  of  35,  54,  56,  63  ? 

3.  What  are  the  factors  of  45,  72,  64,  81,  96? 

96.  Some  numbers  may  be  resolved  into  more  than  two  fac- 
tors; and  also  into  different  sets  of  factors.     Thus,  12  =  2X2X3; 
also  12  =  4X3  =  6X2. 

4.  What  are  the  different  factors  and  sets  of  factors  of  8,  16, 
18,  20,  24? 

5.  What  are  the  different  factors  and  sets  of  factors  of  27,  32, 
86,  40,  48? 

9G.  a.  We  have  seen  that  the  product  of  any  two  numbers  is 
the  same,  whichever  factor  is  taken  for  the  multiplier.  (Art.  83.) 
In  like  manner,  it  may  be  shown  that  the  product  of  any  three  01 

QUKST.— 95.  What  is  a  composite  number  ?  Ob?.  What  are  the  factors  which  produce  it 
sometimes  called  ?  What  is  meant  by  resolving  a  number  into  factors?  l)fi.  Are  numbers 
ever  composed  of  more  thnn  two  factors?  96.  a.  When  three  or  more  factors  are  to  be 
multiplied  together,  does  *  make  any  difference  in  what  order  they  ate  taken  ? 


ARTS.  94-97.J  MULTIPLICATION.  57 

more  factors  will  be  the  same,  in  whatever  order  they  are  multi- 
plied. For,  the  product  of  two  factors  may  be  considered  as  one 
number,  and  this  may  be  taken  either  for  the  multiplicand,  or  the 
multiplier.  Again,  the  product  of  three  factors  may  be  consid- 
ered as  one  number,  and  be  taken  for  the  multiplicand,  or  the  mul- 
tiplier, &c.  Thus,  24  =  3X2X2X2  =  6X2X2  =  12X2  =  6X4  = 
4X2X3  =  8X3. 

CASE  1. —  When  the  multiplier  is  a  composite  number. 

6.  What  will  27  bureaus  cost,  at  31  dollars  apiece? 
Analysis. — Since  27  is  three  times  as  much  as  9  ;  that  is,  27  =  9 

X3,  it  is  manifest  that  27  bureaus  will  cost  three  times  as  much 
as  9  bureaus. 

Operation. 

Dolls.  31  cost  of  1  B.  Having  resolved  27  into  the  factors 

9  9  and  3,  we  find  the  cost  of  9  bureaus, 

Dolls.  279  cost  of  9  B.  then  multiplying  that  by  3,  we  have 

3  the  cost  of  27  bureaus. 
Dolls.  "837  cost  of  27  B. 

7.  What  will  36  oxen  cost,  at  43  dollars  per  head  ? 
Solution. — 36  =  9X4;  and  43X9X4  =  1548  dolls.  Ans. 

Or,  36  =  3X3X4;  and  43 X 3X3X4= 1548 dolls.  Ans.  Hence, 

97.  To  multiply  by  a  composite  number. 

Resolve  the  multiplier  into  two  or  more  factors ;  multiply  the 
multiplicand  by  one  of  these  factors,  and  this  product  by  another 
factor,  and  so  on  till  you  have  multiplied  by  all  the  factors.  The 
last  product  ivill  be  the  answer  required. 

OBS:  The  factors  into  which  a  number  may  be  resolved,  must  not  be  con- 
founded with  the  parts  into  which  it  may  be  separated.  (Art.  53.)  The  former 
have  reference  to  multiplication,  the  latter  to  addition ;  that  is,  factors  must  be 
multiplied  together,  but  parts  must  be  added  together  to  produce  the  given 
number.  Thus,  56  may  be  resolved  into  two  factors,  8  and  7 ;  it  may  be  sep- 
arated into  two  parts,  5  tens  or  50,  and  6.  Now,  8X!7=56,  and  50-j-6=56. 

8.  What  will  24  horses  cost,  at  74  dollars  a  head  ? 

QUEST. — 97.  When  the  multiplier  is  a  composite  number,  how  do  you  proceed  ?  Obs 
What  is  the  difference  between  the  factors  into  which  a  number  may  be  resolved  and  the 
parts  into  which  it  may  be  separated  7 


58  MULTIPLICATION.  |  SfiCT    IV, 

9.  What  cost  45  hogsheads  of  tobacco,  at  128  dollars  a  hogs 
head? 

10.  What  cost  54  acres  of  land,  at  150  dollars  per  acre? 

11.  At  118  shillings  per  week,  how  much  will  it  cost  a  family 
to  board  49  weeks  ? 

12.  If  a  man  travels  at  the  rate  of  372  miles  a  day,  how  far 
will  he  travel  in  64  days  ? 

13.  At  163  dollars  per  ton,  how  much  will  72  tons  of  lead  cost  ? 

14.  What  cost  81  pieces  of  broadcloth,  at  245  shillings  apiece? 

15.  What  cost  84  carriages,  at  384  dollars  apiece? 

CASE  II. —  When  t/ie  multiplier  is  1  with  ciphers  annexed  to  it. 

98*  It  is  a  fundamental  principle  of  notation,  that  each  re 
moval  of  a  figure  one  place  towards  the  left,  increases  its  value 
ten  times;  (Art.  36;)  consequently,  annexing  &  cipher  to  a  number 
will  increase  its  value  ten  times,  or  multiply  it  by  10;  annexing 
two  ciphers  will  increase  its  value  a  hundred  times,  or  multiply  it 
by  100 ;  annexing  three  ciphers  will  increase  it  a  thousand  times, 
or  multiply  it  by  1000,  &c.  Thus,  15  with  a  cipher  annexed,  be- 
comes 150,  and  is  the  same  as  15X10;  15  with  two  ciphers  an- 
nexed, becomes  1500,  and  is  the  same  as  15  X 100  ;  15  with  thre? 
ciphers  annexed,  becomes  15000,  and  is  the  same  as  15X1000, 
&c.  Hence, 

99.  To  multiply  by  10,  100,  1000,  &c. 

Annex  as  many  ciphers  to  the  multiplicand  as  there  are  ciplien 
in  the  multiplier,  and  the  number  thus^  formed  will  bi  the  product 
required. 

Note. — To  annex  means  to  place  after,  or  at  the  right  hand. 

16.  What  will  ten  boxes  of  lemons  cost,  at  63  shillings  per 
box?     Ans.  630  shillings. 

17.  How  many  bushels  of  corn  will  465  acres  of  land  produce, 
at  100  bushels  per  acre? 

QUEST. — 98.  What  is  the  effect  of  annexing  a  cipher  to  a  number?  Two  ciphers  1 
Three  ?  Four?  99.  How  do  you  proceed  when  the  multiplier  is  10, 100,  1000,  &c.  1  JVot* 
What  Is  the  meaning  of  the  term  annex  ? 


ARTS.  98-100  ]  MULTIPLICATION.  59 

18  Allowing  365  days  for  a  year,  how  many  days  are  there  in 
1000  yeais? 

19.  Multiply  153486  by  10000. 

20.  Multiply  3120467  by  100000. 

21.  Multiply  52690078  by  1000000. 

22.  Multiply  689063457  by  10000000. 

23.  Multiply  4946030506  by  100000000. 

24.  Multiply  87831206507  by  1000000000. 

25.  Multiply  67856005109  by  10000000000. 

CASE  Ill.—  WJien  the  multiplier  has  ciphers  on  the  right  hana. 

26.  What  will  30  wagons  cost,  at  45  dollars  apiece  ? 

Note. — Any  number  with  ciphers  on  its  right  hand,  is  obviously  a  composite 
number;  the  significant  figure  or  figures  being  one  factor,  and  1,  with  thl 
given  ciphers  annexed  to  it,  the  other  factor.  Thus,  30  may  be  resolved  into 
the  factors  3  and  10.  We  may  therefore  first  multiply  by  3  and  then  by  10, 
by  annexing  a  cipher  as  above. 

Solution. — 45X3  =  135,  and  135X10=1350  dolls.  Ans. 

27.  How  many  acres  of  land  are  there  hi  3000  farms,  if  each 
farm  contains  475  acres  ? 

Analysis. — 3000=3  X 1000.  Now  475  X  Operation. 

3  =  1425  ;  and  adding  three  ciphers  to  this  475 

product,  multiplies  it  by  1000.  (Art.  99.)  3 

Hence,  Ans.  1425000  acres. 

1 OO.  When  there  are  ciphers  on  the  right  of  the  multiplier. 

Multiply  the  multiplicand  by  the  significant  figures  of  the  multi- 
plier, and  to  this  product  annex  as  many  ciphers,  as  are  found  on  the 
right  of  the  multiplier. 

OBS.  It  will  be  perceived  that  this  case  combines  the  principles  of  the  two 
preceding  cases ;  for,  the  multiplier  is  a  composite  number,  and  one  of  its  fac- 
tors is  *  with  cipliers  annexed  to  it. 

28  How  much  will  50  hogs  weigh,  at  375  pounds  apiece? 

29  If  1   barrel  of  flour  weighs   192  pounds,  how  much  will 
600  barrels  weigh  ? 

30.  Multiply  14376  by  25000. 

QUEST.— 100.  When  there  are  ciphers  on  the  right  of  the  multiplier,  how  do  you  pro 
oeed  1  Obs.  What  principles  does  this  case  combine  1 


60  MULTIPLICATION.  [SECT.  IV 

31.  Multiply  350634  by  410000. 

32.  Multiply  4630425  by  6200000. 

CASE  IV. —  When  the  multiplicand  has  ciphers  on  the  right  hand, 

33.  What  .will  37  ships  cost,  at  29000  dollars  apiece? 
Analysis.— 29000  =  29X1000.    But  the  Operation. 

product  of  two  or  more  factors  is  the  same  29000 

in   whatever   order   they  are   multiplied.  37 

(Art   96.  a.)     We  therefore  multiply  29  203 

by  37,  and  this  product  by  1000  by  adding  87 

three  ciphers  to  it.  Ans.  1073000    dolls. 

PROOF. — 29000X37=1073000,  the  same  as  before.     Hence, 
1 0 1  •  When  there  are  ciphers  on  the  right  of  the  multiplicand. 
Multiply  the  significant  figures  of  the  multiplicand  by  the  mul- 
tiplier, and  to  the  product  annex  as  many  ciphers,  as  are  found  on 
the  right  of  the  multiplicand. 

OBS.  When  both  the  multiplier  and  multiplicand  have  ciphers  on  the  right, 
multiply  the  significant  figures  together  as  if  there  were  no  ciphers,  and  to  their 
product  annex  as  many  ciphers,  as  are  found  on  the  right  of  both  factors 

34.  Multiply  2370000  by  52. 

35.  Multiply  48120000  by  48. 

36.  Multiply  356300000  by  74. 

37.  Multiply  1623000000  by  89. 

38.  Multiply  540000  by  700. 

Analysis. — 540000=54X10000,    and  Operation. 

700=7  X 1 00  ;  we  therefore  multiply  the  540000 

significant  figures,  or  the  factors  54  and  7 00 

7  together,  (Art.  96.  a,)  and  to  this  pro-  Ans.  378000000 

duct  annex  six  ciphers.  (Art.  99.) 

39.  Multiply  1563800  by  20000. 

40.  Multiply  31230000  by  120000. 

41.  Multiply  5310200  by  3400000. 

42.  Multiply  82065000  by  8100000. 

43.  Multiply  210909000  by  5100000. 

QUEST.— 101.  When  there  are  ciphers  on  the  right  of  the  multiplicand,  how  proceed  1 
Oft*.  How,  when  there  are  ciphers  on  the  right  both  of  the  multiplier  and  multiplicand? 


ARTS.  101-104."]          MULTIPLICATION.  61 

102.  There  are  oilier  methods  of  contracting  the  operations  in 
multiplication,  which,  in  certain  cases,  may  be  resorted  to  with 
advantage.     Some  of  the  most  useful  are  the  following. 

44.  How  many  gallons  of  water  will  a  hydrant  discharge  in  13 
hou^s,  if  it  discharges  2325  gallons  per  hour  ? 

Operation.  Multiplying  by  the  3  units,  we  set  the 

2325  X 13  first  figure  of  the  product  one  place  to  ths 

6975  right  of  the  multiplicand.     Now,  since 

Ant.  30225  gallons,  multiplying  by  1  is  taking  the  multipli- 
cand onct,  (Art.  82,)  we  add  together  the  multiplicand  and  tho 
partial  product  already  obtained,  and  the  result  is  the  answer. 

PROOF. — 2325X13=30225  gallons,  the  same  as  above.     Hence, 

103.  To  multiply  by  13,  14,  15,  &c.,  or  1,  with  either  of  the 
other  digits  annexed  to  it. 

Multiply  by  the  units'  figure  of  the  multiplier,  and  write  each 
figure  of  the  partial  product  one  place  to  the  right  of  that  from 
which  it  arises  ;  finally,  add  the  partial  product  io  the  multipli- 
cand, and  the  result  will  be  the  answer  required. 

Note. — This  method  is  the  same,  in  effect,  as  if  we  actually  multiplied  by  the 
1  ten,  and  placed  the  first  figure  of  the  partial  product  under  the  figure  by 
which  we  multiply.  (Art.  87.  II.) 

45.  Multiply  3251  by  14.  46,  Multiply  4028  by  17. 
47.  Multiply  25039  by  16.  48.  Multiply  50389  by  18. 
49.  If  21  men  can  do  a  job  of  work  in  365  days,  how  long 

will  it  take  1  man  to  do  it  ? 

Operation.  We  first  multiply  by  the  2  tens,  and  set 

365X21  the  first  product  figure  in  tens'  place,  then 

730  adding  this  partial  product  to  the  multipli- 

Ans.  7665  days,  cand,  we  have  7665,  for  the  answer. 

PROOF.— 365X21  =  7665  days,  the  same  as  above.     H(nce, 

1.O4.  To  multiply  by  21,  31,  41,  &c.,  or  1  with  either  <&f  the 
other  significant  figures  prefixed  to  it. 

Multiply  by  the  tens'  figure  of  the  multiplier,  and  write  the  first 

0 


62  MULTIPLICATION.  [SfiCT.  IV. 

figure  of  the  partial  product  in  tens'  place  /  finally,  add  this  par- 
tial product  to  the  multiplicand,  and  the  result  will  be  the  answer 
required. 

Note. — The  reason  of  this  method  of  contraction  is  substantially  the  same 
as  that  of  the  preceding. 

50    Multiply  4275  by  31.  51.  Multiply  7504  by  41. 

52.  Multiply  38256  by  61.  53.  Multiply  70267  by  81, 

54.  How  much  will  99  carriages  cost,  at  235  dollars  apiece  ? 

Analysis. — Since  1  carriage  costs  235  Operation. 

dollars,  100  carriages  will  cost  100  times  23500  price  of  100  C 
as  much,  which  is  23500  dollars.  (Art.  235  "  of  1  C. 
99.)  But  we  wished  to  find  the  cost  23265  "  of  99  C. 
of  99  carnages  only.  Now  99  is  1  less 

than  100 ;  therefore,  if  we  subtract  the  price  of  1  carriage  from 
the  price  of  100,  it  will  give  the  price  of  99  carriages.  Hence, 

1O5.  To  multiply  by  9,  99,  999,  or  any  number  of  9s. 

Annex  as  many  ciphers  to  the  multiplicand  as  there  are  9s  in  the 
multiplier  ;  from,  the  result  subtract  tJie  given  multiplicand,  and 
tJie  remainder  will  be  the  answer  required. 

Note. — The  reason  of  this  method  is  obvious  from  the  fact  that  annexing  as 
many  ciphers  to  the  multiplicand  as  there  are  9s  in  the  multiplier,  multiplies  it 
by  100,  or  repeats  it  once  more  than  is  required;  (Art.  99;)  consequently,  sub- 
tracting the  multiplicand  from  the  number  thus  produced,  must  give  the  true 
answer, 

55.  Multiply  4791  by  99.  56.  Multiply  6034  by  999. 
57.  Multiply  7301  by  999.  58.  Multiply  463  by  9999. 
59.  What  is  the  product  of  867  multiplied  by  84  ? 

Analysis. — We  first  multiply  by  4  in  the  usual          Operation. 
way.     Now,  since  8=4  X  2,  it  is  plain,  if  the  par-  867 

tial  product  of  4  is  multiplied  by  2,  it  will  give  84 

the  partial  product  of  8,     But  as  8  denotes  tens,  3468X2 

the  first  figure  of  its  product  will  also  be  tens.         6936 
(Art.  86.)     The  sum  of  the  two  partial  products         72328  Ans. 
will  be  the  answer  required. 

Note. — For  the  sake  of  convenience  in  multiplying,  the  factor  2  is  placed  at 
the  right  of  the  partial  product  of  4,  with  the  sign  X>  between  them. 


ARTS.  105,  106.]          MULTIPLICATION.  (33 

60.  What  is  the  product  of  987  by  486  ? 

Operation. 

987  Since  48  =  6  X  8,  we  multiply  the  partial  prod- 

486  uct  of  6  by  8,  and  set  the  first  product  figure 

5922X8  in  tens' place  as  before.  (Art.  86.) 

47376 

479682  Ans. 

PROOF. — 987x486  =  479682,  the  same  as  above.     Hence, 

1  O6«  When  part  of  the  multiplier  is  a  composite  number  of 
which  the  other  figure  is  a  factor. 

First  multiply  by  the  fiyure  that  is  a  factor  ;  then  multiply  this 
partial  product  by  the  other  factor,  or  factors,  taking  care  to  write 
the  first  figure  of  each  frirtial  product  in  its  proper  order,  and 
their  sum  will  be  tht  ansit^r  required.  (Art.  86.) 

OBS.  When  the  figure  m  thousands,  ten  thousands,  or  any  other  column,  is 
a  factor  of  the  other  par*,  or  )  arts  of  the  multiplier,  care  must  be  taken  to 
place  the  first  figure  of  its  product  under  the  factor  itself,  and  the  first  figure 
of  each  of  the  other  partial  products  in  its  own  order.  (Art.  86.) 

(61.)  (62.) 

2378  256841 

936  85632 

21402   X4  2054728    7x4 

__85608  14383096 

2225808  Ans.  8218912 

21993808512  Ans. 

63.  Multiply  665  by  82.  64.  Multiply  783  by  93. 

65.  Multiply  876  by  396.          66.  Multiply  69412  by  95436, 
67.  324325X54426.  68.  256721X85632. 

69.  What  is  the  product  of  63  multiplied  by  45  ? 
No1e. — By  multiplying  the  figures  which  produce  the  same  order,  and  add- 
ing the  results  mentally,  we  may  obtain  th*  '-nswer  without  setting  down  the 
partial  products. 

First,  multiplying  the  units  into  units,  we  se 

Operation.         down  the  result  and  carry  as  usual.     JSTow,  since 

63  the  6  tens  into  5  «mits,  and  3  units  into  4  tens  will 

45  both  produce  th«  same  order,  viz :  .tens,  (Art.  86,) 

2835  Ans.     we  multiply  them  and  add  their  products  men- 


64 


MULTIPLICATION. 


[SECT.  IV. 


tally.     Thus,  6X5  =  30,  and  3  X  4  =  1 2  ;  now,  30+  1 2  =  42,  and  1 
(to  carry)  makes  43.     Finally,  6X4  =  24,  and  4  (to  carry)  make  28. 

PROOF. — 63X45  =  2835,  the  same  as  before.     Hence, 

1O7»  To  multiply  any  two  numbers  together  without  setting 
down  the  partial  products. 

First  multiply  the  units  together  /  then  multiply  the  figures 
'(?"' ih  produce  tens,  and  adding  the  products  mentally,  set  down  the 
tsult  and  carry  as  usual.  Next  multiply  the  figures  which  produce 
Hundreds,  and  add  the  products,  &c.,  as  before.  In  like  manner, 
perform  the  multiplications  which  produce  thousands,  ten  thou- 
sands, (fcc.,  adding  tJie  products  of  each  order  as  you  proceed,  and 
thus  continue  the  (peration  till  all  the  figures  are  multiplied. 

70.  What  is  the  product  of  12346789  into  54321  ? 

Analytic  Operation. 


23456789 
54321 

2X5 

2X4 
3X5 

2X3 
3X4 
4X5 

2X2 
3X3 
4X4 
5X5 

2X1 
3X2 
4X3 
5X4 
6X5 

3X1 
4X2 
5X3 
6X4 
7X5 

4X1 
5X2 
6X3 
7X4 
8X5 

5X1 

6X2 
7X3 
8X4 
9X5 

6X1 
7X2 
8X3 
9X4 

7X1 

8X2 
9X3 

8X1 

9X2 

9X1 

0         6 


6        9 


Explanation. — Having  multiplied  by  the  first  two  figures  of  the 
multiplier,  as  in  the  last  example,  we  perceive  that  there  are  three 
.multiplications  which  will  produce  hundreds,  viz :  7  X  1>  8X2,  and 
9X3  ;  (Art.  86  ;)  we  therefore  perform  these  multiplications,  add 
their  products  mentally,  and  proceed  to  the  next  order.  Again, 
there  are  four  multiplications  which  will  produce  thousands,  viz: 
6  X  1,  7  X  2,  8  X  3,  and  9  X  4.  (Art.  86.)  We  perform  these  mul- 
tiplications as  before,  and  proceed  in  a  similar  manner  through  all 
the  remaining  orders.  Ans.  706296235269. 

Note. — 1.  In  the  solution  above,  the  multiplications  of  the  different,  figures  aro 
arranged  in  separate  columns,  that  the  various  combinations  which  produce 
the  same  order,  may  be  seen  at  a  glance.  In  practice  it  is  unnecessary  to  de- 
note these  multiplications.  The  principle  being  understood,  the  process  of 


ARTS.  107,  108.)          MULTIPLICATION. 


65 


multiplying  and  adding  may  easily  be  carried  on  in  the  mind,  while  the  final 
product  only  is  set  down. 

2.  When  the  factors  contain  but  two  or  three  figures  each,  this  method 
is  very  simple  and  expeditious.  A  little  practice  will  enable  the  student  to 
apply  it  with  facility  when  the  factors  contain  six  or  eight  figures  each,  and 
its  application  will  afford  an  excellent  discipline  to  the  mind.  It  has  sometimes 
been  used  when  the  factors  contain  twenty- four  figures  each;  but  it  is  doubt-^. 
ful  whether  the  attempt  to  extend  it  so  far,  'is  profitable. 


71,  Multiply  25X25. 

73  Multiply  81X64. 

75.  Multiply  194X144. 

77.  Multiply  4825X2352. 


72.  Multiply  54X54. 
74.  Multiply  45X92. 
76.  Multiply  1234X125. 
78.  Multiply  6521X5312. 


1O8.  By  suitable  attention,  the  critical  student  will  discovei 
various  other  methods  of  abbreviating  the  processes  of  multipli- 
cation. 

Solve  the  following  examples,  contracting  the  operations  when 
practicable. 

79.  42634X63. 

80.  50035X56. 

81.  72156X1000. 

82.  42000X40000. 

83.  80000X25000. 

84.  2567345X17. 

85.  4300450X19. 

86.  9803404X41, 

87.  6710045X71 

88.  3456710X18 

89.  7000541X91. 

90.  4102034X99. 

91.  42304X999. 

92.  50421X9999. 

93.  67243X99999. 

94.  78563X93. 

95.  34054X639. 

96.  52156X756. 

97.  41907X54486 

98.  26397X24648. 


99.  12900X14000. 

100.  64172X42432. 

101.  26815678X81. 

102.  85X85. 

103.  256X256. 

104.  322X325. 

105.  5234X2435. 

106.  48743000X637. 

107.  31890420X85672. 

108.  80460000X2763. 

109.  2364793X8485672. 

110.  1256702X999999. 
6840005X91X61. 
45067034X17X51, 
788031245X81X16. 
61800000X23000. 
12563000X4800000. 
91300203X1000000. 

117.  680040000X1000000. 

118.  4000000000X1000000, 


111. 
112. 
113. 
114. 
115. 
116. 


66  DIVISION.  [SECT.  V 

SECTION     V. 

DIVISION. 

ART.  11O.  Ex.  1.  How  many  barrels  of  flour,  at  8  dollar 
per  barrel,  can  you  buy  for  56  dollars  ? 

Analysis. — Since  flour  is  8  dollars  a  barrel,  it  is  obvious  you 
can  buy  1  barrel  as  often  as  8  dollars  are  contained  in  56  dollars; 
and  8  dolls,  are  contained  in  56  dolls.  7  times.  Ans.  7  barrels. 

Ex.  2.  A  man  wished  to  divide  72  dollars  equally  among  9  beg- 
gars :  how  many  dollars  would  each  receive  ? 

Solution. — Reasoning  as  before,  each  beggar  would  receive  as 
many  dollars  as  9  is  contained  times  in  72 ;  and  9  is  contained  in 
72,  8  times.  Ans.  8  dollars. 

OBS.  The  learner  will  at  once  perceive  that  the  object  in  the  first  example, 
is  to  find  kow  many  times  one  number 'is  contained  in  another;  and  that  the 
object  of  the  second,  is  to  divide  a  given  number  into  equal  parts,  but  its  solu- 
tion consists  in  finding  how  many  times  one  number  is  contained  in  another, 
and  is  the  same  in  principle  as  that  of  the  first. 

Ill*  The  Process  of  finding  how  many  times  one  number  is 
contained  in  another,  is  called  DIVISION. 

The  number  to  be  divided,  is  called  the  dividend. 

The  number  by  which  we  divide,  is  called  the  divisor. 

The  number  obtained  by  division,  or  the  answer  to  the  question, 
is  called  the  quotient.  It  shows  how  many  times  the  divisor  is 
contained  in  the  dividend.  Hence,  it  may  be  said, 

112*  Division  is  finding  a  quotient,  which  multiplied  into  the 
divisor,  will  produce  the  dividend. 

Note. — The  term  quotient  is  derived  from  the  Latin  word  guoties,  which  sig- 
nifies how  often,  or  how  many  times. 

QUEST.— 111.  What  is  division  ?  What  is  the  number  to  be  divided  called  1  The  num- 
ber by  which  we  divide  ?  What  is  the  number  obtained  called  ?  What  does  the  quotient 
show  ?  112.  What  then  may  division  be  said  to  be  ? 


ARTS.  110-114.]  DIVISION.  67 

113.  The  number  which  is  sometimes  left  after  division,  is 
called  the  remainder.     Thus,  when  we  say  5  is  contained  in  38, 
7  times,  and  3  over,  5  is  the  divisor,  38  the  dividend,  7  the  quo- 
tient, and  3  the  remainder. 

OBS.  1.  The  remainder  is  of  the  same  denomination  as  the  dividend;  for,  It 
IN  a  part  of  it. 

2.  The  remainder  is  always  less  than  the  divisor ;  for,  if  it  were  equzl  to, 
or  greater  than  the  divisor,  the  divisor  could  be  contained  once  more  !n  the 
dividend. 

114.  It  will  be  perceived  that  division  is  similar  in  principle 
to  subtraction,  and  may  be  performed  by  it.     For  instance,  to  tind 
how  many  times  7  is  contained  in  21,  subtract  7  (the  divisor)  con- 
tinually from  2 1  (the  dividend),  until  the  latter  is  exhausted  ;  then 
counting  these  repeated  subtractions,  we  shall  have  the  true  quo- 
tient.    Thus,  7  from  21  leaves  14  ;  7  from  14  leaves  7  ;  and  7  from 
7  leaves  0.     Now  by  counting,  we  find  that  7  has  been  taken  from 
21,  3  times  ;  consequently,  7  is  contained  in  21,  3  times.     Hence, 

Division  is  sometimes  defined  to  be  a  short  way  of  performing 
repeated  subtractions  of  the  same  number. 

OBS.  1.  It  will  be  observed  that  division  is  the  reverse  of  multiplication. 
Multiplication  is  the  repeated  addition  of  the  same  number;  division  is  the 
repeated  subtraction  of  the  same  number.  The  product  of  the  one  answers  to 
the  dividend  of  the  other ;  but  the  latter  is  always  given,  while  the  former  is 
required. 

2.  When  the  dividend  denotes  things  of  one  denomination  only,  the  opera 
tion  is  called  Simple  Division. 

SHORT   DIVISION. 

Ex.  3.  How  many  hats,  at  2  dollars  apiece,  can  be  bought  for 
4862  dollars? 

Operation.  We  write  the  divisor  on  the  left  of  the  divi- 

Divisor.   Divid.  dend  with  a  curve  line  between  them;  then, 

beginning  at  the  right  hand,  proceed  thus  :  2  is 
Quot.  2431  contained  in  4,  2  times.     Now,  since  the  4  de 

QUEST.— 113.  What  is  the  number  called  which  is  sometimes  left&fter  division  1  Cbs.  Of 
what  denomination  is  the  remainder?  Why?  Is  the  remainder  greater  or  less  than  the 
divisor?  Why?  114.  Tfc  what  rule  is  division  similar  in  principle?  Obs.  Of  what  is 
division  the  reverse  ?  When  the  dividend  denotes  things  of  one  denomination  only,  what 
ts  the  operation  called  1 

T.H.  4 


68  DIVISION.  [SECT.  V 

notes  thousands,  the  2  must  be  thousands ;  we  therefore  wnte  it 
in  thousands'  place,  under  the  figure  divided.  2  is  contained  in 
8,  4  times ;  and  as  the  8  is  hundreds,  the  4  must  also  he  hun- 
dreds; hence  we  write  it  in  hundreds'  place,  under  the  figure 
divided.  2  in  6,  3  times ;  the  6  being  tens,  the  3  must  also  be 
tens,  and  should  be  set  in  tens'  place.  2  in  2,  once ;  and  since 
the  2  is  units,  the  1  is  a  unit,  and  must  therefore  be  written  in 
inits'  place.  The  answer  is  2431  hats. 

1 1 5  •  When  the  process  of  dividing  is  carried  on  in  the  mind, 
and  the  quotient  only  is  set  down,  as  in  the  last  example,  the  opera- 
tion is  called  SHORT  DIVISION. 

116.  The  reason  that  each  quotient  figure  is  of  the  same  order 
AS  the  figure  divided,  may  be  shown  in  the  following  manner : 

Having  separated  the  dividend 

Analytic  Solution.  of  the  last  example  into  the  orders 

4862  =  4000  +  800  +  60  +  2       of  which  it  is  composed,  we  per- 

2)4000  +  800  +  60  +  2       ceive  that  2  is  contained  in  4000, 

2000+400  +  30  +  1       2000  times;    for  2X2000  =  4000, 

Again,  2  is  contained  in  800,  400 

times;  for  2X400  =  800,  &c.  Am.  2431. 

Ex.  4.  A  man  left  an  estate  of  209635  dollars,  to  be  divided 
equally  among  4  children :  how  much  did  each  receive  ? 

Since  the  divisor  4,  is  not  contained  in 

Operation.  2,  the  first  figure  of  the  dividend,  we  find 

4)209635  how  many  times  it  is  contained  in  the  first 

Ans.      52408-f  dolls.      two  figures.     Thus,  4  is  contained  in  20, 

5  times  ;  write  the  5  under  the  0.    Again, 

4  is  contained  in  9,  2  times  and  1  over ;  set  the  2  under  the  9. 
Now,  as  we  have  1  thousand  over,  we  prefix  it  mentally  to  the 
6  hundreds,  making  16  hundreds;  and  4  in  16,  4  times.  Write 
the  4  under  the  6.  But  4  is  not  contained  in  3,  the  next  figure, 
we  therefore  put  a  cipher  in  the  quotient,  and  prefix  the  3  to  the 
next  figftre  of  the  dividend,  as  if  it  were  a.  remainder.  Then  4  ia 
35,  8  times  and  3  over ;  place  the  8  under  the  5,  and  setting  the  re- 
mainder over  the  divisor  thus  f,  place  it  on  the  right  of  the  quotient, 
Note. — To  prefix  means  to  place  before,  or  at  the  left  hand. 


ARTS.  115-118.]  DIVISION.  69 

117.  When  the  divisor  is  not  contained  in  any  figure  of  the 
dividend,  a  cipher  must  always  be  placed  in  the  quotient. 

OBS.  The  reason  for  placing  a  cipher  in  the  quotient,  is  to  preserve  the  true 
local  value  of  each  figure  of  the  quotient.  (Art.  116.) 

118.  In  order  to  render  the  division  complete,  it  is  obvious 
that  the  whole  of  the  dividend  must  be  divided.     But  when  there 
is  a  remainder  after  dividing  the  last  figure  of  the  dividend,  it 
must  of  necessity  be  smaller  than  the  divisor,  and  cannot  be  di- 
vided by  it.  (Art.  113.  Obs.  2.)     We  therefore  represent  the  divi- 
sion by  placing  the  remainder  over  the  divisor,  and  annex  it  to 
the  quotient.  (Art.  28.) 

OBS.  1.  The  learner  will  observe  that  in  dividing  we  begin  at  the  left  hand, 
instead  of  the  right,  as  in  Addition,  Subtraction,  and  Multiplication.  The  rea- 
son is,  because  there  is  frequently  a  remainder  in  dividing  a  higher  order, 
which  must  necessarily  be  united  with  the  next  lower  order,  before  the  division 
can  be  performed. 

2.  The  divisor  is  placed  on  the  left  of  the  dividend,  and  the  quotient  under 
it,  merely  for  the  sake  of  convenience.  When  division  is  represented  by  the 
sign  -f-,  the  divisor  is  placed  on  the  right  of  the  dividend :  and  when  repre- 
sented in  the  form  of  a  fraction,  the  divisor  is  placed  under  the  dividend, 

LONG  DIVISION. 

Ex.  5.  At  15  dollars  apiece,  how  many  cows  can  be  bought 
for  3525  dollars? 

Operation. 
Having  written  the  divisor  on  the  left  of          Divisor.  Divid.   Quot. 

the  dividend  as  before,  we  find  that  15  is  15)  3525  (235 

contained  in  35,  2  times,  and  place  the  2  on  30 

the  right  of  the  dividend,  with  a  curve  line  52 

between  them.     We  next  multiply  the  di-  45 

visor  by  this  quotient  figure,  place  the  prod-  75 

uct  under  the  figures  divided,  and  subtract  75 

it  therefrom.     We  ruw  bring  down  the  next 

figure  of  the  dividend,  and  placing  it  on  the  right  of  the  remainder 
5,  we  perceive  that  15  is  contained  in  52,  3  times.  Set  the  3  on 
the  right  of  the  last  quotient  figure,  multiply  the  divisor  by  it,  and 
subtract  the  product  from  the  figures  divided  as  before.  We  then 


70  DIVISION."  [SECT.  V 

briii**1  down  the  next,  which  is  the  last  figure  of  the  dividend,  to 
the  right  oi  this  remainder,  and  finding  15  is  contained  in  75, 
5  times,  we  place  the  5  in  the  quotient,  multiply  and  subtract  as 
before.  The  answer  is  235  cows. 

119.  When  the  result  of  each  step  in  the  operation  is  written 
down,  as  in  the  last  example,  the  process  is  called  LONG  DIVISION. 
Long  Division  is  the  same  in  principle  as  Short  Division.  The 
only  difference  between  them  is,  that  in  the  former,  the  result  of 
each  step  in  the  operation  is  written  down,  while  in  the  latter, 
we  carry  on  the  process  in  the  mind,  and  simply  write  the  quotient. 

OBS.  1.  When  the  divisor  contains  but  one  figure,  the  operation  by  Slwrt 
Division  is  the  most  expeditious,  and  therefore  should  always  be  practiced ; 
but  when  the  divisor  contains  two  or  more  figures,  it  will  generally  be  the  most 
convenient  to  use  Long  Division. 

2.  To  prevent  mistakes,  it  is  advisable  to  put  a  dot  under  each  figure  of  the 
dividend,  when  it  is  brought  down. 

3.  The  French  place  the  divisor  on  the  right  of  the  dividend,  and  the  quo- 
tient below  the  divisor,*  as  seen  in  the  following  example. 

Ex.  6.  How  many  times  is  72  contained  in  5904  ? 

Operation. 

5904  (72  divisor.  The  divisor  is  contained  in  590,  the 

576      82  quotient.  first  three  figures  of  the  dividend,  8 

144  times.     Set  the  8  under  the  divisor, 

144  multiply,  &c.,  as  before. 

Ex.  7    How  many  times  is  435  contained  in  262534  ? 

Operation.  Since  the  divisor  is  not  contained 

A35)262534(603fff  Am.  in  the  first  three  figures  of  the  divi- 

2610  '"'  dend,  we  find  how  many  times  it  is 

1534  contained  in  the  first  four,  the  few- 

1305  est  that  will  contain  it,  and  write 

229  rem.  the  6  in  the  quotient ;  then  multi- 

VjrasT.  -115.  What  is  short  division  ?  119.  What  is  long  division  ?  What  is  the  dif- 
itttou\.e  Uotween  them  1 

*  Elements  D'Ari  thine  tique,  par  M.  Bourdon.  Also,  Lacroix's  Arithmetic,  translated  by 
Pri/e*$oi  Farrar. 


ARTS.  119,  120.]  DIVISION.  71 

plying  and  subtracting  as  before,  the  remainder  is  15.  Bringing 
down  the  next  figure,  we  have  153  to  be  divided  by  435.  Bui 
435  is  not  contained  in  153  ;  we  therefore  place  a  cipher  in  tin 
quotient,  and  bring  down  the  next  figure.  Then  435  in  1534,  9 
times.  Place  the  3  in  the  quotient,  and  proceed  as  before. 

Note. — After  the  first  quotient  figure  is  obtained,  for  each  figure  of  the  ami 
dend  which  is  brought  down,  either  a  significant  figure,  or  a  cipher,  must  be  pu. 
in  the  quotient.  (Art.  116.) 

1 2O«  From  the  preceding  illustrations  and  principles  we  de- 
rive the  following 

GENERAL   RULE   FOR   DIVISION. 

I.  When  the  divisor  contains  but  one  figure. 

Write  tlie  divisor  on  tlie  left  of  the  dividend,  with  a  curve  lint 
between  them.  Begin  at  the  left  hand,  divide  successively  each 
figure  of  the  dividend  by  the  divisor,  and  place  each  quotient  figure 
directly  under  the  figure  divided.  (Arts.  116, 118.  Obs.  1,  2.) 

If  there  is  a  remainder  after  dividing  any  figure,  prefix  it  to 
the  next  figure  of  the  dividend  and  divide  this  number  as  before  ; 
and  if  the  divisor  is  not  contained  in  any  figure  of  the  dividend, 
place  a  ciplier  in  the  quotient  and  prefix  this  figure  to  the  next 
one  of  the  dividend,  as  if  it  were  a  remainder.  (Arts.  117,  118.) 

II.  When  the  divisor  contains  more  than  one  figure. 
Beginning  on  the  left  of  tJie  dividend,  find  how  many  times  the 

divisor  is  contained  in  the  fewest  figures  that  will  contain  it,  and 
place  the  quotient  figure  on  the  right  of  the  dividend  with  a  curve 
line  between  them.  Tlicn  multiply  tlie  divisor  by  this  figure  and 
subtract  tlie  product  from  the  figures  divided  ;  to  the  right  of  tlte 
remainder  bring  down  the  next  figure  of  the  dividend  and  divide 
this  number  as  before.  Proceed  in  this  manner  till  all  tJie  figure! 
of  the  dividend  are  divided. 

QUEST.— 12f .  How  do  you  write  the  numbers  for  division  "?  When  the  divisor  contelni 
but  one  figure,  how  proceed  ?  Why  place  the  divisor  on  the  left  of  the  dividend  and  th« 
quotient  unde  the  figure  divided?  When  there  is  a  remainder  after  dividing  a  fig  ire, 
what  is  to  be  done  with  it  ?  When  the  divisor  is  not  contained  in  any  figure  of  the  divl 
dend  how  proceed  ?  Why  ?  Why  begin  to  divide  at  the  left  hand  ?  When  the  divtaoi 
contains  more  than  one  figure,  how  proceed  1 


72  DIVISION  [SECT.  V^. 

Wlienever  there  is  a  remainder  after  dividing  the  last  figure, 
write  it  over  the  divisor  and  annex  it  to  the  quotient.  (Art.  118.) 

Demonstration.  —  The  principle  on  which  the  operations  in  Division  depend, 
Is  that  a  part  of  the  quotient  is  found,  and  the  product  of  this  part  into  the 
divisor  is  taken  from  the  dividend,  showing  how  much  of  the  latter  remains  to 
be  divided;  then  another  part  of  the  quotient  is  found,  and  its  product  into  the 
divisor  is  taken  from  what  remained  before.  Thus  the  operation  proceeds  till 
the  whole  of  the  dividend  is  divided,  or  till  the  remainder  is  less  than  the  divisot 
(Art.  113.  Obs.2.) 

.  OBS  When  the  divisor  is  large,  the  pupil  will  find  assistance  in  determining 
the  quotient  figure,  by  finding  how  many  times  the  first  figure  of  the  divisor  iat 
contained  in  the  first  figure,  or  if  necessary,  the  first  two  figures  of  the  divi- 
dend. This  will  give  pretty  nearly  the  right  figure.  Some  allowance  must, 
however,  be  made  for  carrying  from  the  product  of  the  other  figures  of  the  di- 
visor, to  the  product  of  the  first  into  the  quotient  figure. 

121*  PROOF.  —  Multiply  the  divisor  by  the  quotient,  to  the 
product  add  the  remainder,  and  if  the  sum  is  equal  to  the  dividend, 
the  work  is  right. 

OBS.  Since  the  quotient  shows  how  many  times  the  divisor  is  contained  in 
the  dividend,  (Art.  Ill,)  it  follows,  that  if  the  divisor  is  repeated  as  many  timeg 
as  there  are  units  in  the  quotient,  it  must  produce  the  dividend. 

Ex.  8.  Divide  256329  by  723. 

Operation.  Proof. 

723)256329(354fff  Ans.  723  divisor. 

2169  354  quotient. 

3942  2892 

3615  3615 

3279  2169 

2892  387  rem. 

387  rem.  256329  dividend. 

122*  Second  Method.  —  Subtract  the  remainder,  if  any,  from 
the  dividend,  divide  the  dividend  thus  diminished,  by  the  quotient 
and  if  the  result  is  equal  to  the  given  divisor,  the  work  is  right. 


When  there  is  a  remainder  after  dividing  the  last  figure  of  the  dividend,  what 
must  be  done  with  it?  121.  How  is  division  proved  ?  Obs.  How  does  it  appear  that  the 
product  of  the  divisor  and  quotient  will  be  equal  to  the  dividend,  if  the  work  Js  right  1 
Can  division  be  proved  by  any  other  methods  ? 


ARTS.  121- 127.  J  DIVISION.  73 

123.  Third  Method.— First  cast  the  9s  out  of  the  divisor 
and  quotient,  and  multiply  the  remainders  together  ;  to  the  prod- 
uct add  the  remainder,  if  any,  after  division ;  cast  the  9s  out  of 
this  sum,  and  set  down  the  excess ;  finally  cast  the  9s  out  of  the 
dividend,  and  if  the  excess  is  the  same  as  that  obtained  from  the 
divisor  and  quotient,  the  work  may  be  considered  right. 

Note. — Since  the  divisor  and  quotient  answer  to  the  multiplier  and  muitipli- 
umd,  and  the  dividend  to  the  product,  it  is  evident  that  the  principle  of  casting 
out  the  9s  will  apply  to  the  proof  of  division,  as  well  as  that  of  multiplication. 
(Art.  90.) 

124.  Fourth  Method. — Add  the  remainder  and  the  respective  products  of 
the  divisor  into  each  quotient  figure  together,  and  if  the  sum  is  equal  to  the 
dividend,  the  work  is  right. 

Note. — This  mode  of  proof  depends  upon  the  principle  that  the  wlwle  of  a 
fuantity  is  equal  to  the  sum  of  all  its  parts.  (Ax.  11.) 

125.  Fifth  Method. — First  cast  the  11s  out  of  the  divisor  and  quotient,  and 
.multiply  the  remainders  together;  to  the  product  add  the  remainder,  if  any, 
after  division,   and  casting  the   lls  out  of  this  sum,  set  down  the  excess; 
dually,  cast  the  lls  out  of  the  dividend,  and  if  the  excess  is  the  same  as  that 
obtained  from  the  divisor  and  quotient,  the  work  is  right.     (Art.  92.  Note  2.) 

EXAMPLES    FOR    PRACTICE. 

127,  Ex.  1.  A  farmer  raised  2975  bushels  of  wheat  on  45 
acres  of  land  :  how  many  bushels  did  he  raise  per  acre  ? 

2.  A  garrison  consumed   8925   barrels  of  flour  in  105  days  : 
how  much  was  that  per  day  ? 

3.  The  President  of  the  United  States  receives  a  salary  of  25000 
dollars  a  year  :  how  much  is  that  per  day  ? 

4.  A  drover  paid  2685  dollars  for  895  head  of  cattle  :   how 
much  did  he  pay  per  head  ? 

5.  If  a  man's  expenses  are  3560  dollars  a  year,  how  much  are 
they  per  week  ? 

6.  If  the  annual  expenses  of  the  government  are  27  millions 
of  dollars,  how  much  will  they  be  per  day  ? 

7.  How  long  will  it  take  a  ship  to  sail  from  New  York  U 
Liverpool,  allowing  the  distance  to  be  3000  miles,  and  the  ship 
to  sail  144  miles  per  day  ? 

8.  Sailing  at  the  same  rate,  how  long  would  it  take  the  same 
ship  to  sail  round  the  globe,  a  distance  of  25000  miles  ? 

7 


74 


DIVISION. 


[SECT.  V 


10.  47839  —  42. 

11.  75043-52. 

12.  93840  —  63. 

13,  421645- 

-74. 

14  325000- 

-85. 

15,  400000- 

-96. 

16.  999999- 

-47. 

17.  352417- 

-29. 

18.  47081-7- 

251. 

19.  423405- 

-485. 

20.  16512-7- 

344. 

21.  304916- 

-6274. 

22.  12689-7- 

145. 

23.  145260- 

-1345. 

24.  147735- 

-3283. 

25.  1203033  —  327. 

26.  1912500—425. 

27.  5184673  —  102. 

28.  301140-7-478. 

29.  8893810—37846. 

30.  9302688—14356. 

31.  9749320—365. 

32.  3228242  —  5734. 

33.  75843639426-7-8593. 

34.  65358547823^-2789. 

35.  102030405060-7-123456. 

36.  908070605040-7-654321. 

37.  1000000000000000—111. 

38.  1000000000000000—1111. 

39.  1000000000000000—11111 


CONTRACTIONS   IN  DIVISION. 

128.  The  operations  in  division,  as  well  as  those  in  multipli- 
cation, may  often  be  shortened  by  a  careful  attention  to  the  appli- 
cation of  the  preceding  principles. 

CASE  1. —  When  the  divisor  is  a  composite  number. 

Ex.  1.  A  man  divided  837  dollars  equally  among  27  persons, 
who  belonged  to  3  familiet,  each  family  containing  9  persons : 
tow  many  dollars  did  each  person  receive  ? 

Analysis. — Since  27  persons  received  837  dollars,  each  one 
must  have  received  as  many  dollars,  as  27  is  contained  times  in 
837.  But  as  27  (the  number  of  persons),  is  a  composite  number 
whose  factors  are  3  (the  number  of  families),  and  9  (the  number 
of  persons  in  each  family),  it  is  obvious  we  may  first  find  how 
many  dollars  each  family  received,  and  then  how  many  each  per- 
son received 


Operation. 

3)837  whole  sum  divided. 

9)279  portion  of  each  Fam. 
Ans.  31       "        "      "     person. 


If  3  families  received  337 
dollars,  1  family  must  have 
received  as  many  dollars,  as 
3  is  contained  times  ir  837 


and  3  in  837,  279  times.    That  is,  each  family  received  279  dollar! 


ARTS.  128,  129.]  DIVISION.  "75 

Again,  if  9  persons,  (the  number  in  each  family,)  received  279  dol- 
lars, 1  person  must  have  received  as  many  dollars,  as  9  is  con- 
tained times  in  279  ;  and  9  in  279,  31  times.     Ans.  31  dollars. 
PROOF.—  -31X27=837,  the  same  as  the  dividend.     Hence, 


To  divide  by  a  composite  number. 

1.  Divide  the  dividend  by  one  of  the  factors  of  the  divisor,  then 
iivids  the  quotient  thus  obtained  by  another  factor  ;  and  so  on  till 
ill  the  factors  are  employed.     The  last  quotient  will  be  the  answer. 

II.  To  find  the  true  remainder. 

If  the  divisor  is  resolved  into  but  two  factors,  multiply  the  last 
remainder  by  the  first  divisor,  to  the  product  add  the  first  remain- 
der, if  any,  and  the  result  will  be  the  true  remainder. 

When  more  than  two  factors  are  employed,  multiply  each  re- 
mainder by  all  the  preceding  divisors,  to  the  sum  of  their  prod- 
ucts, add  the  Jirst  remainder,  and  the  result  will  be  the  true  re- 
mainder. 

OBS.  1.  The  true  remainder  may  also  be  found  by  multiplying  the  quotient 
by  the  divisor,  and  subtracting  the  product  from  the  dividend. 

2.  This  contraction  is  exactly  the  reverse  of  that  in  multiplication.    (Art  97.) 
The  result  will  evidently  be  the  same,  in  whatever  order  the  factors  are  taken. 

2.  A  man  bought  a  quantity  of  clover  seed  amounting  to  507 
pints,  which  he  washed  to  divide  into  parcels  containing  64  pints 
each  :  how  rnary  parcels  can  lie  make  ? 

Note.  —  Since  64=2x8X4,  we  divide  by  the  factors  respectively 
Operation. 
2)507 

8)253~—  1  rem.       ...         =    1  pt. 
4)31  —  5  rem.     Now  5X2         =10  pts. 
7—3  rem.     and  3X8X2     =  48  pts. 
Ans.  7  parcels,  and  59  pts.  over.  59  pts.  True  Rem. 

Demonstration,  —  1.  Dividing  507  the  number  of  pints,  by  2,  gives  253  fo  .to 

quotient,  or  distributes  the  seed  into  253  equal  parcels,  leaving  1  pint  t  AT 

Now  the  units  of  this  quotient  are  evidently  of  a  different  valiie  from  thos  jf 

the  given  dividend  ;  for  since  there  are  but  half  as  many  parcels  as  at  fir  it 

QUBST.—  129.  How  proceed  when  the,  divisor  is  a  composite  number]  How  flu  he 
true  remainder  1 

4* 


76  DIVISION.  [SECT.  V. 

is  plain  that  each  parcel  must  contain  2  pints,  or  1  quart;  that  is,  every  unit 
of  the  fi  .-st  quotient  contains  2  of  the  units  of  the  given  dividend  ;  consequently, 
every  unit  of  it  that  remains  will  contain  the  same  ;  (Art.  113.  Obs.  2  ;)  there- 
fore this  remainder  must  be  multiplied  by  2,  in  order  to  find  the  units  of  the 
given  dividend  which  it  contains. 

2.  Dividing  the  quotient  253  parcels,  by  8,  will  distribute  them  into  31  other 
equal  parcels,  each  of  which  will  evidently  contain  8  times  the  quantity  of  the 
preceding,  viz  :  8  times  1  quart  =8  quarts,  or  1  peck;  that  is,  every  unit  of  th 
second  quotient  contains  8  of  the  units  in  the  first  quotient,  or  8  times  2  of  th 
units  in  the  given  dividend  ;  therefore  what  remains  of  it,  must  be  multiplied 
by  8x2,  or  16,  to  find  the  units  of  the  given  dividend  which  it  contains. 

3.  In  like  manner,  it  may  be  shown,  that  dividing  by  each  successive  factor 
reduces  each  quotient  to  a  class  of  units  of  a  higher  value  than  the  preced- 
ing ;  that  every  unit  which  remains  of  any  quotient,  is  of  the  same  value  as 
that  quotient,  and  must  therefore  be  multiplied  by  all  the  preceding  divisors,  in 
order  to  find  the  units  of  the  given  dividend  which  it  contains. 

4.  Finally,  the  several  remainders  being  reduced  to  the  same  units  as  those 
of  the  given  dividend  according  to~  the  rule,  their  sum  must  evidently  be  the 
true  remainder.     (Ax.  11.) 

3.  How  many  acres  of  land,  at  35  dollars  an  acre,  can  you  buy 
for  4650  dollars  ? 

4.  Divide  16128  by  24.  5.  Divide  25760  by  56. 

6.  Divide  17220  by  84.  7.  Divide  91080  by  72. 

•* 

CASE  II.  —  When  the  divisor  is  1  with  cipliers  annexed  to  it. 

13O«  It  has  been  shown  that  annexing  a  cipher  to  a  number 
increases  its  value  ten  times,  or  multiplies  it  by  10.  (Art.  98.) 
Reversing  this  process  ;  that  is,  removing  a  cipher  from  the  right 
hand  of  a  number,  will  evidently  diminish  its  value  ten  times,  or 
divide  it  by  10  ;  for,  each  figure  in  the  number  is  thus  restDred 
to  its  original  place,  and  consequently  to  its  original  value.  1  hus, 
annexing  a  cipher  to  15,  it  becomes  150,  which  is  the  same  as 
15X10.  On  the  oth?r  hand,  removing  the  cipher  from  150,  it 
becomes  15,  which  is  the  same  as  150-f-10. 

In  the  same  manner  it  may  be  shown,  that  removing  two  ciphers 
from  the  right  of  a  number,  divides  it  by  100;  removing  three,  di- 
vides it  by  1000  ;  removing  four,  divides  it  by  10000,  (fee.  Hence, 


—130.  What  is  the  effect  of  annexing  a  cipher  to  a  number  ?    What  is  the  effect 
of  removing  a  apher  from  the  right  of  a  number  1    How  does  this  appear  1 


ARTS.  130-132.]  DIVISION.  77 

131.  To  divide  by  10,  100,  1000,  &c. 

Cut  off  as  *many  figures  from  the  right  hand  of  the  dividend  as 
tliere  are  ciphers  in  the  divisor.  The  remaining  figures  of  tlie  div- 
idend will  lie  the  quotient,  and  those  cut  off  the  remainder. 

d.  In  one  dime  there  are  10  cents  :  how  many  dimes  are  there 
in  200  cents  ?  In  340  cents  ?  In  560  cents  ? 

9.  In  one  dollar  there  are   100   cents  :  how  many  dollars  are 
there  in  C5000  cents  ?     In  765000  cents  ?     In  4320000  cents  9 

10.  Divide  26750000  by  100000. 

11.  Divide  144360791  by  1000000. 

12.  Divide  582367180309  by  100000000. 

CASE  III. —  When  the  divisor  has  cipliers  on  tlie  right  Jiand. 

13.  How  many  hogsheads  of  molasses,  at  30  dollars  apiece, 
can  you  buy  for  9643  dollars  ? 

OBS.  The  divisor  30,  is  a  composite  numl>er,  the  factors  of  which  are  3  and 
10.  (Arts.  05, 9(5.)  We  may,  therefore,  divide  first  by  one  factor  and  the 
quotient  thence  arising  by  the  other.  (Art.  121). )  Now  cutting  off  the  right 
hand  figure  of  the  dividend,  divides  it  by  ten  ;  (Art.  131 ;)  consequently  divid- 
ing the  -remaining  figures  of  the  dividend  by  3,  the  other  factor  of  the  divisor, 
Will  give  the  quotient. 

Operation.  We  first  cut  off  the  cipher  on  the  right 

3|0)964|3  of  the  divisor,  and  also  cut  off  the  right 

321  -J-jj-  Ans.        hand  figure  of  the  dividend  ;  then  divid- 
ing  964    by    3,  we   have    1    remainder. 

l»ow  as  the  3  cut  off,  is  part  of  the  remainder,  we  therefore 
««nex  it  to  the  1.  Ans.  '321-J-ft  hogsheads.  Hence, 

1 32.  When  there  are  ciphers  on  the  right  hand  of  the  divisor. 
Cut  off  tlie  ciphers,  also  cut  off  as  many  figures  from  the  right 

of  the  dividend.  Then  divide  the  other  figures  of  the  dividend  by 
ihe  remaining  figures  of  the  divisor,  and  annex  tlie  figures  cut  off 
from  the  dividend  to  tlie  remainder. 

14.  How  many  buggies,  at  70  dollars  apiece,  can  you  buy  foi 
7350  dollars  ? 

QUEST.— 131.  How  proceed  when  the  divisor  is  10,  100,  1000,  &c.  ?  ir2.  When  there  art 
tiph«rs  on  the  right  hand  of  the  divisor,  how  proceed  1  What  is  to  bo  done  with  figure* 
cut  off  from  the  dividend  ? 


78  DIVISION. 


[SECT.  V. 


15.  How  many  barrels  will  it  take  tc   pack  36800  pounds  of 
pork,  allowing  200  pounds  to  a  barrel  ? 

16.  Divide  3360000  by  17000. 

133.  Operations  in  Long  Division  may  be  shortened  by  sub- 
tracting the  product  of  the  respective  figures  in  the  divisor  into 
each  quotient  figure  as  we  proceed  in  the  operation,  setting  down 
the  remainders  only.  This  is  called  the  Italian  Method. 

17.  How  many  times  is  21  contained  in  4998  ? 

Operation. 

21)4998(238         This  method,  it  will  be  seen,  requires  a  much 

79  smaller  number  of  figures  than   the  ordinary 

168  process. 

18.  Divide  1188  by  33.  19.  Divide  2516  by  37. 
20.  Divide  3128  by  86.  21.  Divide  7125  by  95. 

22.  A  merchant  laid  out  873  dollars  in  flour,  at  5  dollars  a 
barrel :  how  many  barrels  did  he  get  ? 

Operation.  We  first  double  the  dividend,  and  then  di- 

873  vide  the  product  by  10,  which  is  done  by 

2  cutting  off  the  right  hand  figure.    (Art.  131.) 

110)174|6  But  since  we  multiplied  the  dividend  by  2,  it 

174  f  Ans.     is  plain  that  the  6  cut  off,  is  2  times  too  large 

for  the  remainder ;  we  therefore  divide  it  by 

2,  and  we  have  8  for  the  true  remainder.     Hence, 

134*  When  the  divisor  is  5. 

Multiply  the  dividend  by  2,  and  divide  the  product  ~hy  10. 
(Art.  131.)" 

tfote. — 1.  When  the  figure  cut  off  is  a  significant  figure,  it  must  be  divided 
by  2  for  the  true  remainder. 

2.  This  contraction  depends  upon  the  princvpto  that  any  given  divisor  ig 
contained  in  any  given  dividend,  just  as  man^.  times  as  twice  that  divisor  is 
contained  in  twice  that  dividend,  three  times  that  divisor  in  three  times  that  divi- 
dend, &c.  For  a  further  illustration  of  this  principle  see  General  Pr  pciplcs 
m  Division 

23.  Divide  6035  by  5.  24.  Divide  8450  by  5. 
25.  Divide  32561  by  5.  26.  Divide  43270  ly  5. 


ARTS.  133-139.]  DIVISION.  79 

135*  When  the  divisor  is  15,  35,  45,  or  55. 

Double  the  dividend,  and  divide  t/ie  product  by  30,  70,  90,  or 
110,  as  tlie  case  ma    be.  (Art.  132.) 

Note. — This  method  is  simply  doubling  both  the  divisor  and  dividend.  V\  o 
must  there  tore  divide  the  remainder,  if  any,  by  2,  for  the  true  remainder. 

'   27.  Divide  3256  by  15.  28.  Divide  2673  by  35. 

29.  Divide  3507  by  45.  30.  Divide  7853  by  55 

136.  When  the  divisor  is  25. 

Multiply  tJie  dividend  by  4,  and  divide  the  product  by  100 
(Art.  131.) 

Note. — This  is  obviously  the  same  as  multiplying  both  the  dividend  and  divi- 
sor by  4.  (Art.  134.  Note  2.)  Hence,  we  must  divide  the  remainder,  if  any 
"thus  found,  by  4,  for  the  true  remainder. 

31.  Divide  2350  by  25.  32.  Divide  4860  by  25. 

33.  Divide  42340  by  25.  34.  Divide  94880  by  25. 

137.  To  divide  by  125. 

Multiply  the  dividend  by  8,  and  divide  the  product  by  1000. 
(Art.  131.) 

Note. — This  contraction  is  multiplying  both  the  dividend  and  divisor  by  8. 
For  the  true  remainder,  therefore,  we  must  divide  the  remainder,  if  any,  by  8. 

35.  Divide  8375  by  125.  36.  Divide  25426  by  125. 

138.  To  divide  by  75,  175,  225,  or  275. 

Multiply  the  dividend  by  4,  and  divide  the  product  by  300,  700 
900,  or  1100,  as  the  case  may  be.   (Art.  132.) 
Note. — For  the  t^ue  remainder,  divide  the  remainder,  if  any  thus  found,  by  4 
37.  Divide  1125  by  75.  38.  Divide  2876  by  175. 

39.  Divide  3825  by  225.  40.  Divide  8250  by  275. 

139.  The  preceding  are  among  the  most  frequent  and  useful 
modes    of    contracting    operations   in   division.     Various   other 
methods  might  be  added,  but  they  will  naturally  suggest  them- 
selves to  the  inventive  student,  as  opportunities  occur  for  their 
application. 

41.  How  long  would  it  take  a  vessel  sailing  100  miles  per  .lav 
to  cirtumnavigate  the  earth,  whose  circumference  is  25000  miles? 


80  DIVISION.  [SECT.  V. 

42.  The  distance  of  the  Earth  from  the  Sun  is  95,000,000  of 
miles  :  how  long  would  it  take  a  balloon  going  at  the  rate  of 
100,000  miles  a  year,  to  reach  the  sun  ? 

43.  The  debts  of  the  several  States  of  the  Union,  in  1840, 
amounted  to  171,000,000  of  "dollars,  and  the  number  of  inhabi- 
tants was  17,000,000  :  How  much  must  each  individual  have  beeUr 
taxed  to  pay  the  debt? 

44.  The  national  debt  of  Holland  is  800,000,000  of  dollars, 
And  the  number  of  inhabitants  2,800,000  :  what  is  the  amount 
of  indebtedness  of  each  individual  ? 

45.  The  national  debt  of  Spain  is  467,000,000  of  dollars,  and 
the  number  of  inhabitants   11,900,000:  what  is  the  amount  of 
indebtedness  of  each  individual  ? 

46.  The  national  debt  of  Russia  is  150,000,000  of  dollars,  and 
the  number  of  inhabitants  51,100,000  :  what  is  the  amount  of 
indebtedness  of  each  individual  ? 

47.  The  national  debt  of  Austria  is  380,000,000  of  dollars, 
and  the  number  of  inhabitants  34,100,000  :  what  is  the  amount 
of  indebtedness  of  each  individual  ? 

48.  The  national  debt  of  France  is  1,800,000,000  of  dollars, 
and  the  number  of  inhabitants  33,300,000 :  what  is  the  amount 
of  indebtedness  of  each  individual  ? 

49.  The  national  debt  of  Great  Britain  is  5,556,000,000   of 
dollars,  and  the  number  of  inhabitants  25,300,000  :  what  is  the 
amount  of  indebtedness  of  each  individual  ? 

50.  Divide  467000000000  by  25000000000. 

51.  568240—42.  62.  462156  —  75. 

52.  785372  —  63.     x  63.   3562189^-225. 
'53.  896736  —  72.  64.   685726  —  32000 

54.  67234568H-5.  65.   723564—175. 

55.  34256726-M5.  66.   892565  —  225. 

56.  42367581^-45,  67.  456212  —  275. 

57.  16753672  —  35.  68.  925673  —  125. 

58.  3256385-7-55.  69.  763421  —  175. 

59.  45672400^25.  70.  876240  —  275. 
dO,  6245634-^45.  71.  7825600-^-80000. 

'•*  8245623—125  72.  9200-1578-:- 100000. . 


Airis.  139-143.]  DIVISION.  81 

GENERAL   PRINCIPLES  'iN   DIVISION. 

'.  4O.  From  the  nature  of  division,  it  is  evident,  that  the 
value  of  the  quotient  depends  both  on  the  divisor  and  the  divi- 
dend. 

141.  If  a  given  divisor  ia  contained  in  a  given  dividend  a 
certain  number  of  times,  the  same  divisor  will  obviously  be  con- 
tained, 

In  double  that  dividend,  twice  as  many  times. 

In  three  times  that  dividend,  thrice  as  many  times,  <fec.     Hence, 

If  the  divisor  remains  the  same,  multiplying  the  dividend  by  any 
number,  is  in  effect  multiplying  the  quotient  by  that  number. 

Thus,  6  is  contained  in  12,  2  times  ;  in  2  times  12  or  24,  6  is 
contained  4  times  ;  (i.  e.  twice  2  times  ;)  in  3  times  12  or  36,  6 
is  contained  6  times  ;  (i.  e.  thrice  2  times  ;)  &c. 

142.  Again,  if  a  given  divisor  is  contained  in  a  given  divi- 
dend a  certain  number  of  times,  the  same  divisor  is  contained, 

In  half  that  dividend,  half  as  many  times  ; 

In  a  third  of  that  dividend,  a  third  as  many  times,  &c.     Hence, 

If  the  divisor  remains  the  same,  dividing  the  dividend  by  any 
number,  is  in  effect  dividing  the  quotient  by  that  number. 

Thus,  8  is  contained  in  48,  6  times  ;  in  48-7-2  or  24,  (half  of 
48,)  8  is  contained  3  times ;  (i.  e.  half  of  6  times  ;)  in  48-7-3  or 
10,  (a  third  of  48,)  8  is  contained  2  times  ;  (i.  e.  a  third  of  6 
times ;)  &c. 

143.  If  a  given  divisor  is  contained  in  a  given  dividend  a 
certain  number  of  times,  then,  in  the  same  dividend, 

Twice  that  divisor  is  contained  only  half  as  many  tunes ; 
Three  times  that  divisor,  a  third  as  many  times,  &c.     Hence, 
If  tlie  dividend  remains  the  same,  multiplying  tlie  divisor  by  any 
number,  is  in  effect  dividing  the  quotient  by  that  number. 

Thus,  4  is  contained  in  24,  6  times  ;  2  times  4  or  8  is  ccn 

QUEST.— 140.  Upon  what  does  the  value  of  the  quotient  depend  1  141.  If  the  divisor  rr 
mains  the  same,  what  is  the  effect  on  the  quotient  to  multiply  the  dividend?  142.  Wh* 
is  the  effect  of  dividing  the  dividend  by  any  given  number  7  143.  If  the  divide  id  rema:: 
the  same,  what  is  the  effect  of  multiplying  the  divisor  by  any  given  number  1 


82  DIVISION.  [SE*-T.  V. 

tained  in  24,  3  times  ;  (i.  e.  half  of  6  times  ;)  3  times  4  01  12  is 
contained  in  24,  2  times  ;  (i.  e.  a  third  of  6  times  ;)  &o. 

144*  If  a  given  divisor  is  contained  in  a  given  divicVad  a 
certain  number  of  times,  then,  in  the  same  dividend, 

Half  that  divisor  is  contained  twice  as  many  times  ; 

A  third  of  that  divisor,  three  times  as  many  times,  &c.     H^nce, 

[f  the  dividend  remains  the  same,  dividing  the  divisor  by  any 
tdmber,  is  in  effect  multiplying  the  quotient  by  tliat  number. 

Thus,  6  is  contained  in  36,  6  times  ;  6-7-  2  or  3,  (half  of  6,)  is 
contained  in  36,  12  times;  (i.  e.  twice  6  times;)  .6-r-3  or  2,  (a 
third  of  6,)  is  contained  in  36,  18  times  ;  (i.  e.  thrice  6  times  ;)  &c. 

145*  From  the  preceding  articles,  it  is  evident  that  any  given 
divisor  is  contained  in  any  given  dividend,  just  as  many  times  as 
twice  that  divisor  is  contained  in  twice  that  dividend  ;  three  times 
that  divisor  in  three  times  that  dividend,  &c. 

Conversely,  any  given  divisor  is  contained  in  any  given  dividend 
just  as  many  times,  as  half  that  divisor  is  contained  in  half  that 
dividend  ;  a  third  of  that  divisor,  in  a  third  of  that  dividend,  <3cc. 
Hence, 

1  46*  If  the  divisor  and  dividend  are  both  multiplied,  or  both 
divided  by  the  same  number,  the  quotient  will  not  be  altered. 

Thus,  6  is  contained  in  12,  2  times  ; 

2  times  6  is  contained  in  2  times  1  2,  2  times  ; 

3  times  6  is  contained  in  3  times  12,  2  times,  &c. 
Again,  12  is  contained  in  48,  4  times  ; 

12-r2  is  contained  in  48-1-2,  4  times  ; 
12-f-3  is  contained  in  48-r-3,  4  times  ;  &c. 

1  47  .  If  the  sum  of  two  or  more  numbers  is  divided  by  Any 
number,  the  quotient  will  be  equal  to  the  sum  of  tlie  quotients 
which  will  arise  from  dividing  the  given  numbers  separately. 

Thus,  the  sum  of  12+18  =  30  ;  and  30-1-6  =  5. 
Now,  12-r6  =  2;  and  18-7-6  =  3;  but  the  sum  of  2  +  3=  5. 

Again,  the  sum  of  32  +  24+40=96  ;  and  96-7-8  =  )  2. 
Now,  32-7-8=4;  24-7-8  =  3;  and  40-7-8=5;  but  4+  3+5=  12 


144.  What  of  dividing  the  divisor?     146.  What  is  the  effect  m  .  i  the  quotient 
If  the  divisor  and  dividend  are  both  multiplied,  or  both  divided  by  the  sai.Mi  number? 


\RTS.   144-150.  J  CANCELATION.  83 

CANCELATION.* 

148.  We  have  seen  that  division  is  finding  a  quotient,  ^hich, 
multiplied  into  the  divisor,  will  produce  the  dividend.  (Art.  .112.) 
If,  therefore,  the  dividend  is  resolved  into  two  such  factors  that 
one  of  them  is  the  divisor,  the  other  factor  will,  of  course,  be  the 
quotient.  Suppose,  for  example,  42  is  to  be  divided  by  6.  Now 
the  factors  of  42  are  6  and  7,  the  first  of  which  being  the  div:«or, 
he  other  must  be  the  quotient.  Therefore, 

Canceling  a  factor  of  any  number,  divides  the  number  by  that 
factor.  Hence, 

1  49.  When  the  dividend  is  the  product  of  two  factors.,  one 
of  which  is  the  same  as  the  divisor. 

Cancel  the  factor  common  to  the  dividend  and  divisor  ;  the 
other  factor  of  the  dividend  will  be  tJie  answer.  (Ax.  9.) 

Note.  —  The  term  cancel,  signifies  to  erase  or  reject. 

1.  Divide  the  product  of  34  into  28  by  34. 
Common  Method.  By  Cancelation. 

34 


28  28  Ans. 

272 

68  Canceling  the  factor  34,  which  is  com- 

34)952(28  Ans.          mon  both  to  the  divisor  and  dividend,  we 
68  have  28  for  the  quotient,  the  same  as  be- 

272  fore. 

272 

15O.   The  method  of  contracting  arithmetical  operations,  by 
rejecting  equal  factors,  is  called  CANCELATION. 

OBS.  It  applies  with  great  advantage  to  that  class  of  examples  and  problems, 
which  involve  both  multiplication  and  division  ;  that  is,  which  require  the  pro- 
duct of  two  or  more  numbers  to  be  divided  by  another  number,  or  by  the  product 
of  two  or  more  numbers. 

2    Divide  76X45  by  76.  3.  Divide  03X81  by  81. 

4.  Divide  65X82  by  82.  5.  Divide  95X73  by  05. 

6.  Divide  the  product  of  45  times  84  by  9. 

*  Birk's  Arithmetical  Collections  :  London,  1764. 


84  APPLICATIONS    OF  [SECT.   V. 


Analysis.  —  The  factor  45  =  5X9>  hence  the  dividend  is  com- 
posed of  the  factors  84X5X9-  We  may  therefore  cancel  9, 
which  is  common  both  to  the  divisor  and  dividend,  and  84  X  5, 
the  )ther  factors  of  the  dividend,  will  be  the  answer  required. 

Operation.  Proof. 

0)81X5X0  84X5X9  =  3780 

420  Ans.  And       3780-7-9  =  420. 

7.  Divide  the  product  of  45  X  6  X  3  by  18  X  5. 

Operation.  Proof. 

*$X5)45X0X#  45X6X3  =  810;  and  18X5  —  90 

9  Ans.  Now,  810-f-90  =  9 

Note.  —  We  cancel  the  factors  6  and  3  in  the  dividend  and  18  in  the  divi- 
scr;  for  GX3=18.  Canceling  the  same  or  equal  factors  in  the  divisor  and 
dividend,  is  dividing  them  both  by  the  same  number,  and  therefore  does  not 
aQ'ect  the  quotient.  (Arts.  14G,  148.)  Hence, 

151.  When  the  divisor  and  dividend  have  common  factors. 

Cancel  the  factors  common  to  both  ;  then  divide  the  product  of 
those  remaining  in  the  dividend  by  the  product  of  those  remaining 
in  the  divisor. 

8.  Divide  15X?X12  by  5X3X?X2. 

9.  Divide  27x3X4x7  by  9X12X6. 

10.  Divide  75X15X24  by  25X3X6X4X5. 

Note.  —  The  further  development  and  application  of  the  principles  of  Cancela- 
tion,  may  be  seen  in  reduction  of  compound  fractions  to  simple  ones;  in  multi- 
plication and  division  of  fractions  ;  in  simple  and  compound  proportion,  &c. 

1  5  1  •  a.  The  four  preceding  rules,  viz  :  Addition,  Subtraction, 

Multiplication,  and  Division,  are  usually  called  the  FUNDAMENTAI 
RULES  of  Arithmetic,  because  they  are  the  foundation  or  basis  or 
all  arithmetical  calculations. 

OBS.  Every  change  that  can  be  made  upon  the  value  of  a  number,  must 
necessarily  either  increase  or  diminish  it.  Hence,  the  fundamental  operation! 
in  arithmetic  are,  strictly  speaking,  but  two,  addition  and  subtraction  ;  that  is 
v&cr-ease  and  decrease.  Multiplication,  we  have  seen,  is  an  abbreviated  form 
of  addition;  division  of  subtraction.  (Arts.  80,  114.) 


151.  a.  Name  the  fundamental  rules  of  Arithmetic.  Why  are  these  rules  called 
fundamental? 


ARTS.  151-154.]      FUNDAMENTAL  RULES.  85 

APPLICATIONS   OF   THE   FUNDAMENTAL   RULES. 

1  52.  When  the  sum  of  two  numbers  and  one  of  the  numbers 
are  given,  to  find  the  other  number. 

From  the  given  sum,  subtract  the,  given  number,  and  the  remainder 
will  be  the  other  number. 

Ex.  1.  The  sum  of  two  numbers  is  87,  one  of  which  is  25: 
what  is  the  other  number  ? 

Solution. — 87" — 25  =  62,  the  other  number.  (Art.  72.) 
PN.OOF. — 62+25  =  87,  the  given  sum.   (Ax.  11.) 

2.  A  and  B  together  own  350  acres  of  land,  95  of  which  be- 
long to  A  :  how  many  does  B  own  ? 

3.  Two  merchants  bought  1785  bushels  of  barley  together,  one 
of  them  took  860  bushels  :  how  many  bushels  did  the  other  have? 

153*  When  the  difference  and  the  greater  of  two  numbers  are 
given,  to  find  the  less. 

Subtract  the  difference  from  the  greater,  and  the  remainder  will 
be  the  less  number. 

4.  The  greater  of  two  numbers  is  72,  and  the  difference  be- 
tween them  is  28  :  what  is  the  less  number  ? 

Solution.— 72 — 28  =  44,  the  less  number.  (Art.  72.) 
PROOF. — 44  +  28  =  72,  the  greater  number.  (Art.  73.  Obs.) 

5.  A  man  bought  a  horse  and  chaise  ;  for  the  chaise  he  gave 
265  dollars,  which  was  75  dollars  more  than  he  paid  for  the 
horse  :  how  much  did  he  give  for  the  horse  ? 

6.  A  traveler  met  two  droves  of  sheep  ;   the  first  contained 
1250,   which  was   125  more  than  the  second  had:  l>ow  many 
Bheep  were  there  in  the  second  drove  ? 

1  5  t.  When  the  difference  and  the  less  of  two  numbers  ar 
given,  to  find  the  greater. 

QUEST. — 152.  When  the  sum  of  two  numbers  and  one  of  them  are  given,  how  Is  the  other 
found  1  153.  When  the  difference  and  the  greater  of  two  numbers  are  {river,  how  is  the 
less  found  ?  154.  When  the  difference  and  the  less  of  two  numbers  are  given,  how  is  the 
jreater  found  I 

8 


86  APPLICATIONS    OP  [SECT.  V 

Add  the  difference  and  the  less  number  together,  and  the  sum  will 
be  tlif.  greater  number.  (Art.  73.  Obs.) 

7.  The  difference  between  two  numbers  is  12,  and  the  less 
number  is  45  :  what  is  the  greater  number  ? 

Solution, — 45+12=57,  the  greater  number. 
PROOF. — 57 — 45=12,  the  given  difference.  (Art.  72.) 

8.  A  is  worth  1890  dollars,  and  B  is  worth  350  dollars  mcc€ 
than  A  :  how  much  is  B  worth  ? 

9.  A  man's  expenses  are  2561  dollars  a  year,  and  his  in-line 
exceeds  his  expenses  875  dollars  :  how  much  is  his  income  \ 

155*  When  the  sum  and  difference  of  two  numbers  are  given, 
to  find  the  two  numbers. 

From  the  sum  subtract  the  difference,  divide  the  remainder  by  2, 
and  the  quotient  will  be  the  smaller  number. 

To  the  smaller  number  thus  found,  add  the  given  difference,  and 
the  sum  will  be  tlie  larger  number. 

10.  The  sum  of  two  numbers  is  48,  and  their  difference  is  18  : 
what  are  the  numbers  ? 

Solution. — 48 — 18=30,  and  30^-2=15,  the  smaller  number. 
And  15  +  18=33,  the  greater  number. 

PROOF. — 33  +  15=48,  the  given  sum.  (Ax.  11.) 

11.  The  sfim  of  the  ages  of  two  men  is  173  years,  and  the 
difference  between  them  is  15  years  :  what  are  their  ages  ? 

12.  A  man  bought  a  span  of  horses  and  a  carnage  for  856 
dollars  ;  the  carriage  was  worth  165  dollars  more  than  the  horses : 
what  was  the  price  of  each  ? 

156.  When  the  product   of  two  numbers  and  one  of  the 

numbers  are  given,  to  find  the  other  number. 

Divide  the  given  product  by  the  given  number,  and  the  p«>tient 
will  be  Hue,  number  required.  (Art.  91.) 

QUEST.— 155.  When  the  sum  and  difference  of  two  numbers  are  given,  how  are  th« 
numbers  found  1  156.  When  the  product  of  two  numbers  and  one  of  them  arc  given,  now 
la  the  other  found  ? 


ARTS.  155-158.]     FUNDAMENTAL  RULES  87 

13.  The  product  of  two  numbers  is  144,  and  one  of  the  nun« 
bers  is  8  :  what  is  the  other  number  ? 

Solution. — 144-r8=18,  the  required  number.  (Art.  120.) 
PROOF. — 18X8  =  144,  the  given  product,  (Art.  88.) 

14.  The  product  of  A  and  B's  ages  is  3250  years,  and  B's  age 
ie  50  years  :  what  is  the  age  of  A  ? 

15.  The  product  of  the  length  of   a  field  multiplied  by  its 
fcieadth  is  15925  rods,  and  its  breadth  is   91  rods  :  what  is  its 
length  ? 

157.  When  the  divisor  and  quotient  are  given,  to  find  the 
dividend. 

Multiply  the  given  divisor  and  quotient  together,  and  the  product 
will  be  the  dividend.  (Art.  121.) 

16.  If  a  certain  divisor  is  12,  and  the  quotient  is  30,  what  is 
the  dividend  ? 

Solution. — 30X12=360,  the  dividend  required. 
PROOF. — 360-^- 12=30,  the  given  quotient.  (Art.  120.) 

17.  If  the  quotient  is  275  and  the  divisor  683,  what  must  be 
the  dividend  ? 

18.  If  the  divisor  is  1031  and  the  quotient  1002,  what  must 
be  the  dividend  ? 

158»  When  the  dividend  and  quotient  are  given,  to  find  the 
divisor. 

Divide  the  given  dividend  by  the  given  quotient,  and  the  quotient 
thus  obtained  will  be  the  number  required.  (Art.  122.) 

19.  A  certain  dividend  is  864,  and  the  quotient  is  12  :  what  is 
the  divisor  ? 

Solution. — 864-^12=72,  the  divisor  required.  (Art.  120.) 
PROOF. — 72X12=864,  the  given  dividend.  (Art.  121.) 

20.  A  gentleman  handed  a  purse  containing  1152  shillings,  to 

QUK-ST.— 157.  When  the  divisor  and  quotient  are  given,  how  is  the  dividend  fccnd  1 
148.  When  the  dividend  and  quotient  are  given,  how  is  the  divisor  found  ? 


88  PROPERTIES  [SECT.  V* 

a  company  of  beggars,  which  was  sufficient  to  give  them  24  shil 
lings  apiece  :  how  many  beggars  were  there  ? 

21.  A  farmer  having  2500  sheep,  divided  them  into  flocks  of 
125  each  :  how  many  flocks  did  they  make  ? 

159.  When  the  product  of  three  numbers  and  two  of  the 

numbers  are  given,  to  find  the  other  number. 

Divide  the  given  product  by  the  product  of  the  two  oiven  w*jn» 
ers,  and  the  quotient  will  be  the  other  number. 

22.  There  are  three  numbers  whose  product  is  288  ;  one  of 
them  is  8  and  another  9  :  it  is  required  to  find  the  other  number. 

Solution. — 9X8=72  ;  and  288-f-72=4,  the  number  required. 
PROOF. — 9X8X4=288,  the  given  product. 

23.  The  product  of  three  persons'  ages  is   14880  years  ;  the 
rtge  of  the  oldest  is  31  years,  and  that  of  the  second  is  24  years  : 
what  is  the  age  of  the  youngest  ? 

24.  If  a  garrison  of   75  men    have   18750  pounds  of  meat, 
how  long  will  it  last  them,  allowing  25  pounds  to  each  man  per 
month  ? 

25.  The  sum  of  two  numbers  is  3471,  and  the  less  is  1629: 
what  is  the  greater  ? 

26.  The  sum  of  two  numbers  is  4136,  and  the  greater  is  3074  : 
what  is  the  less  ? 

27.  The  difference  between  two  numbers  is  128,  and  the  greater 
is  760  :  what  is  the  less  ? 

28.  The  difference  between  two  numbers  is  340,  and  the  less 
is  634  :  what  is  the  greater? 

29.  The  sum  of  two  numbers  is  12640,  and  their  difference  is 
1608  :  what  are  the  numbers  ? 

30.  The  sum  of  two  numbers  is  25264,  and  their  dkTerence 
is  736  :  what  are  the  numbers  ? 

31.  The  sum  of  two  numbers  is  42126,  and  then*  difference  is 
176  :  what  are  the  numbers  ? 

32.  The  product  of  two  numbers  is  246018,  and  one  cr?  them 
is  313  :  what  is  the  other  number  ? 


ARTS.  159,  160."]     PROFER-TIES  OF  NUMBERS.  89 

SECTION    VI. 
PROPERTIES   OF   NUMBERS.* 

ART.  16O.  The  progress  as  well  as  the  pleasure  of  the  student 
i&  Arithmetic,  depends  very  much  upon  the  accuracy  of  his  knowl- 
edge of  the  terms,  which  are  employed  in  mathematical  reasoning. 
Particular  pains  should  therefore  be  taken  to  understand  their 
true  import. 

DBF.  1.  An  integer  signifies  a  whole  number.  (Art.  28.  Obs.  2.) 

2.  Whole  numbers  or  integers  are  divided  into  prime  and  com- 
posite numbers. 

3.  A  composite  number,  we  have  seen,  is  one  which  may  be 
produced  by  multiplying  two  or  more  numbers  together ;  as,  4, 
10,  15.  (Art.  95.) 

4.  A  prime  number  is  one  which  cannot  be  produced  by  multi- 
plying any  two  or  more  numbers  together ;  or  which  cannot  be 
exactly  divided  by  any  whole  number,  except  a  unit  and  itself. 
Thus,  1,  2,  3,  5,  7,  11,  13,  &c.,  are  prime  numbers. 

OBS.  1.  One  number  is  said  to  be  prime  to  another,  when  a  unit  is  the  only 
number  by  which  both  can  be  divided  without  a  remainder. 

2.  The  learnei  must  be  careful  not  to  confound  numbers  which  are  prime 
to  ?ach  other  with  prime  numbers ;  for  numbers  that  are  prime  to  each  other, 
may  thenselves  be  composite  numbers.      Thus  4  and  9  are  prime  to  each 
other,  while  they  are  composite  numbers.  . 

3.  The  number  of  prime  numbers  is  unlimited.     For  those  under  3,  see 
Table,  page  94. 

5.  An  even  number  is  one  which  can  be  divided  by  2  without 
a  remainder ;  as,  4,  6,  8,  10. 


QUEST. — 160.  Upon  what  does  the  progress  and  pleasure  of  the  student  in  Arithmetic 
»ery  much  depend  ?  What  is  an  integer  1  What  is  a  composite  number  ?  What  is  a 
prime  number?  Are  prime  numbers  divisible  by  other  numbers?  Obs.  When  is  one 
number  said  to  be  prime  to  another  ?  How  many  prime  numbers  are  there  ?  What  is  an 
•Ten  number?  An  odd  number?  Obs.  Are  even  numbers  prime  or  composite  ?  What 
Is  tru«  of  odd  numbers  in  this  respect? 

*  Ba  low  on  the  Theory  of  Numbers  ;  also,  Bonny<  istle's  Arithmetic. 


. 

90  PROPERTIES    OF    NUMBERS.  [SECT.  VI. 

6.  An  odd  number  is  one  which  cannot  be  divided  by  2  with- 
out a  remainder ;  as,  1,  3,  5,  7,  0,  15. 

OBS.  All  even  numbers  except  2,  are  composite  numbers ;  an  odd  number  ia 
sometimes  a  composite,  and  sometimes  a  prime  number. 

7.  One  number  is  a  measure  of  another,  when  the  former  is 
contained  in  the  latter,  any  number  of  times  without  a  remainder. 
T&us,  3  is  a  measure  of  15  ;  7  is  a  measure  of  28,  &c. 

8.  One  number  is  a  multiple  of  another,  when  the  former  can 
*b?  divided  by  the  latter  without  a  remainder.     Thus,  6  is  a  mul- 
tiple of  3  ;  20  is  a  multiple  of  5,  &c. 

OBS.  A  multiple  is  therefore  a  composite  number,  and  the  number  thus  con- 
tained in  it,  is  always  one  of  its  factors. 

9.  The  aliquot  parts  of  a  number,  are  the  parts  by  which  it 
can  be  measured  or  divided  without  a  remainder.     Thus,  5  and  7 
are  the  aliquot  parts  of  35. 

10.  The  reciprocal  of  a  number  is  the  quotient  arising  from 
di  viding  a  unit  by  that  number.     Thus,  the  reciprocal  of  2  is  % ; 
the  reciprocal  of  3  is  % ',  &c. 

11.  The  difference  between  a  given  number  and  10,  100,  1000, 
&c.,  that  is,  between  the  given  number  and  the  next  higher  order, 
is  called  the  ARITHMETICAL  COMPLEMENT  of  that  number.     Thus, 
3  is  the  complement  of  7  ;  15  is  the  complement  of  85. 

OBS.  The  arithmetical  complement  of  a  number  consisting  of  one  integral 
figure,  either  with  or  without  decimals,  is  found  by  subtracting  the  number 
from  10.  If  there  are  two  integral  figures,  they  are  subtracted  from  100;  if 
three,  from  1000,  &c. 

12.  A  perfect  number  is  one  which  is  equal  to  the  sum  of  all 
its  aliquot  parts.     Thus,  6  =  1  +  2  +  3,  the  sum  of  its  aliquot  parts, 
and  is  a  perfect  number. 

OBS.  1.  All  the  numbers  known,  to  which  this  property  really  belongs,  are 
the  following:  6;  28;  496;  8128;  33,550,336;  8,589,869,056;  137,438,691,328 
and  2,305,843,008,139,952,128.* 

2.  All  perfect  numbers  terminate  with  6,  or  28. 

QUEST.— When  is  one  number  a  measure  of  another  ?  What  is  a  multiple  »  Wha  laiw 
aliquot  parts  ?  What  is  the  reciprocal  of  a  number  1 

*  Hutton's  Mathematical  Recreations. 


ARTS.  1G),  1GL]       PROPERTIES  OF  NUMBERS.  91 

161.  By  the  term  properties  of  numbers,  is  meant  those 
qualities  or  elements  which  are  inherent  and  inseparable-  frjm 
them.  Some  of  the  more  prominent  are  the  following : 

1.  The  sum  of  any  two  or  more  even  numbers,  is  an  even  number. 

2.  The  difference  of  any  two  even  numbers,  is  an  even  number. 

3.  The  sum  or  difference  of  two  odd  numbers,  is  even  ;  but  the 
sum  of  three  odd  numbers,  is  odd. 

4.  The  sum  of  any  even  number  of  odd  numbers,  is  even ;  but 
the  sum  of  any  odd  number  of  odd  numbers,  is  odd. 

5.  The  sum,  or  difference,  of  an  even  and  an  odd  number,  is  an 
odd  number. 

6.  The  product  of  an  even  and  an  odd  number,  or  of  two  even 
numbers,  is  even. 

7.  If  an   even  number  be  divisible  by  an  odd  number,  the 
quotient  is  an  even  number. 

8.  The  product  of  any  number  of  factors,  is  even,  if  any  one  of 
them  be  even. 

9.  An  odd  number  cannot  be  divided  by  an  even  number  with- 
out a  remainder. 

10.  The  product  of  any  two  or  more  odd  numbers,  is  an  odd 
number. 

31.  If  an  odd  number  divides  an  even  number,  it  will  also 
divide  the  half  of  it. 

12.  If  an  even  number  be  divisible  by  an  odd  number,  it  will 
also  be  divisible  by  double  that  number. 

13.  Any   number   that   measures   two   others,   must   likewise 
measure  their  sum,  their  difference,  and  their  product. 

14.  A  number  that  measures  another,  must  also  measure  its 
multiple,  or  its  product  by  any  whole  number. 

15.  Any  number  expressed  by  the  decimal  notation,  divided 
by  9,  will  leave  the  same  remainder,  as  the  sum  of  its  figures  or 
digits  divided  by  9. 

Demonstration. — Take  any  number,  as  6357 ;  now  separating  it  into  its  seve- 
ral parts,  it  becomes  6000+300+50+7.  But  6000=6 XI 000 =6X(999+1) 
=6X999+6.  In  like  manner  300=3x99+3,  and  50=5X9+5.  Hence 
6357=6X999-(-3x99+5X9+6+3+5+7;  and  6357- 9 =(6x91 9+ 3X99-+ 

Q.UEST. — 161.  What  is  meant  by  properties  <»f  numbers  ? 
T.H.  5 


92  PROPERTIES    OF    NUMBERS.  [SECT.  VL 

5X9+G+3+5+7)-i-9.  But  6X999+3X99+5X9  is  evidently  divisible  by  9 , 
therefore  if  (5357  be  divided  by  9,  it  will  leave  the  same  remainder  as  G-j-3+5-J- 
7-7- !).  The  same  will  be  found  true  of  any  other  .number  whatever. 

OBS.  1.  This  property  of  the  number  9  affords  an  ingenious  method  of  proving 
each  of  the  fundamental  rules.  (Arts.  90,  123.)  The  same  property  belong?  to 
the  number  3;  for,  3  is  a  measure  of  9,  and  will  therefore  be  contained  an  ex- 
o  3t  number  of  times  in  any  number  of  9s.  But  it  belongs  to  no  other  digit. 

2.  The  preceding  is  not  a  necessary  but  an  incidental  property  of  the  num- 

Ler  9.     It  arises  from  the  law  of  increase  in  the  decimal  notation.     If  the  radix 

1  the  system  were  8,  it  would  belong  to  7 ;  if  the  radix  were  12,  it  would  be- 

cwg  to  1 1 ;  and  universally,  it  belongs  to  the  number  that  is  one  less  than  the 

radix  of  the  system  of  notation. 

16.  If  the  number  9  is  multiplied  by  any  single  figure  or  digit 
the  sum  of  the  figures  composing  the  product,   will  make  f. 
Thus,  9X4  =  36,  and  3  +  6  =  9. 

17.  If  we  take  any  two  numbers  whatever ;  then  one  of  them, 
or  their  sum,  or  their  difference,  is  divisible  by  3.     Thus,  take  1 1 
and  1 7  ;  though  neither  of  the  numbers  themselves,  nor  their  sum 
is  divisible  by  3,  yet  their  difference  is,  for  it  is  6. 

18.  Any  number  divided  by  11,  will  leave  the  same  remainder, 
as  the  sum  of  its  alternate  digits  in  the  even  places  reckoning 
from  the  right,  taken  from  the  sum  of  its  alternate  digits  in  the 
odd  places,  increased  by  11  if  necessary. 

Take  any  number,  as  38405603,  and  mark  the  alternate  fig- 
ures-. Now  the  sum  of  those  marked,  viz:  8+0+6  +  3  =  17. 
The  sum  of  the  others,  viz  :  3+4  +  5  +  0=12.  And  17 — 12=5, 
the  remainder  sought.  That  is,  38405603  divided  by  11,  will 
leave  5  remainder. 

Again,  take  5847362,  the  sum  of  the  marked  figures  is  14; 
the  sum  of  those  not  marked  is  21.  Now  21  taken  from  25, 
(—14  +  11,)  leaves  4,  the  remainder  sought. 

19.  Every  composite  number  may  be  resolved  into  prime  factors. 
For,  since  a  composite  number  is  produced  by  multiplying  two  or 
more  factors  together,  (Art.  160.  Def.  3,)  it  may  evidently  be  re- 
solved  into  those  factors  ;  and  if  these  factors  themselves  are  com- 
posite, they  also  may  be  resolved  into  other  factor?,  and  thus  the 
analysis  may  be  continued,  until  all  the  factors  are  prime  numbers. 

20.  The  least  divisor  of  every  number  is  <i  prime  number. 
For,  every  whole  number  is  either  prime,  or  composite  ;  (Art.  160. 


ART.    161.  a.]  PROPERTIES  OF  NUMBERS.  93 

Def.  2  ;)  but  a  composite  number,  we  have  just  seen,  can  be  re- 
solved into  prime  factors  ;  consequently,  the  least  divisor  of  every 
number  must  be  a  prime  number. 

21.  Every  prime  number  except  2,  if  increased  or  diminished 
by  1,  is  divisible  by  4.     See  table  of  prime  numbers,  next  page. 

22.  Every  prime   number   except   2   and   3,  if  increased   or 
diminished  by  1,  is  divisible  by  6. 

23.  Every  prime  number,  except  2  and  5,  is  contained  without 
remainder,  in  the  number  expressed  in  the  common  notation  by 

as  many  9s  as  there  are  units,  less  one,  in  the  prime  number  itself.* 
Thus,  3  is  a  measure  of  99 ;  7  of  999,999 ;  and  13  of  999,999, 
999,999. 

24.  Every  prime  number,  except  2,  3,  and  5,  is  a  measure  of 
the  number  expressed  in  common  notation,  by  as  many  Is  as  there 
are  units,  less  one,  in  the  prime  number.     Thus,  7  is  a  measure 
of  111,111 ;  and  13  of  111,111,111,111. 

25.  All  prime  numbers  except  2,  are  odd ;  and  consequently 
terminate  with  an  odd  digit.  (Art.  160.  Def.  5.) 

Note. — I.  It  must  not  be  inferred  from  this  that  all  odd  numbers  are  prime. 
(Art.  1GO.  Def.  6  Obs.) 

2.  It  is  plain  that  any  number  terminating  with  5,  can  be  divided  by  5  with- 
out a  remainder.  Hence, 

26.  All  prime  numbers,  except  2  and  5,  must  terminate  with 
1,  3,  7,  or  9  ;  all  other  numbers  are  composite. 

1 6 1  •  a.  To  find  the  prime  numbers  in  any  series  of  numbers. 

Write  in  their  proper  order  all  the  odd  numbers  contained  in  the 
series.  Then  reckoning  from  3,  place  a  point  over  every  third  num- 
ber in  the  series  j  reckoning  from  5,  place  a  point  over  every  fifth 
number  •  reckoning  from  7,  place  a  point  over  every  seventh  num- 
ber, and,  so  on.  The  numbers  remaining  without  points,  together 
with  the  number  2,  are  the  primes  required. 

Take  the  series  of  numbers  up  to  40,  thus,  1,  3,  5,  7,  9,  11,  13, 
15,  17,  19,  21,  23,  25,  27,  29,  31,  33,  35,  37,  39  ;  then  adding  tho 
number  2,  the  primes  are  1,  2,  3,  5,  7,  11,  13,  &c. 

AW^.— This  method  of  excluding  the  numbers  which  are  not  prime  from  a 
series,  was  invented  by  Eratosthenes,  and  is  therefore  called  Eratosthenes'  Sieve, 

*  Th£o»ie  des  Nombres,  par  M.  Legendre 


94  PROPERTIES    OF    NUMBERS.  [SECT. 

TABLE    OF    PRIME    NUMBERS    FROM    1    TO    3413. 


11173 

409 

659  941 

12231511 

1811 

2129 

2423  2741 

3079 

2 

179 

419: 

661 

947 

1229 

1523 

1823 

2131 

2437 

2749 

3083 

3 

181 

421 

673  953 

1231 

1531 

1831 

2137 

2441 

2753 

3089 

5 

191 

431 

677 

967 

1237 

1543 

1847 

2141 

2447 

2767 

3109 

7i 

193 

433 

683 

971 

1249 

1549 

1861 

2143 

2459 

2777 

3119 

11 

197 

439 

691 

977 

1259 

1553 

1867 

2153 

2467 

2789 

3121 

13! 

199 

443 

701 

983 

1277 

1559 

1871 

2161 

2473 

2791 

3137 

17 

211 

449 

709 

991 

1279 

1567 

1873 

2179;2477 

2797 

3163 

19 

223 

457 

719 

997 

1283 

1571 

1877 

2203 

25032801 

3167 

23 

227461 

727 

1009 

1289 

15791879 

2207 

2521 

2803 

3169 

29 

229 

463 

733 

1013 

1291 

1583)1889 

2213 

2531 

28193181 

31 

233467J739 

1019 

1297 

1597 

1901 

2221 

2539  2833 

3187 

37 

239 

479 

743 

1021 

1301 

1601 

1907 

2237 

2543 

2837 

3191 

41 

241 

487 

751 

1031 

1303 

1607 

1913 

2239 

2549 

2843  3203 

43 

251 

491 

757 

1033 

1307 

16091931 

2243 

2551 

2851 

3209 

47257 

499 

761 

1039 

1319 

1613 

1933 

2251 

2557 

2857 

3217 

53 

263 

503 

769 

1049 

1321 

1619 

1949 

2267 

2579 

2861 

3221 

59 

269 

509 

773 

1051  1327 

1621 

1951 

226912591 

2879'3229  ' 

61  271 

521 

787 

1061 

1361 

1627 

1973 

2273 

2593 

2887 

3251 

67|277523 

797 

1063 

1367 

1637 

1979 

2281 

2609 

2897 

3253 

71 

281 

541 

809 

1069 

1373 

1657 

1987 

22872617 

2903 

3257 

73 

283 

547 

811 

1087 

1381 

1663  1993 

2293 

2621 

2909 

3259 

79 

293 

557 

821 

1091 

1399 

1667  1997 

2297 

2633 

2917 

3271 

83 

307 

563 

823 

1093 

1409 

1669  1999 

2309  2647 

2927 

3299 

89 

311 

569 

827 

1097 

!1423 

169312003 

2311  2657 

2939 

3301 

97 

313 

571  829 

1103 

J1427 

1697 

2011 

2333 

2659 

2953 

3307 

101 

317 

577  839 

1109 

1429 

1699 

2017 

2339 

2663 

2957 

3313 

103 

331J587 

853 

1117 

1433 

1709 

2027 

2341 

2671 

2963 

3319  ( 

107 

3371593 

857 

1123 

1439 

1721 

2029 

2347 

2677 

2969  3323 

109 

347 

!599 

859 

1129  1447 

1723 

2039 

2351 

2683 

2971 

3329 

113 

349601 

863 

1151 

;1451 

1733 

2053 

2357 

2687 

29993331 

127 

353 

:607 

877 

1153 

1453 

1741 

2063 

2371 

2689 

3001 

13343 

131 

359613 

;881 

1163  1459 

17472069 

2377|2693 

3011 

'3347 

137 

367 

;ei7 

883 

1171 

1471 

1753 

2081 

2381 

2699  3019 

13359 

139 

373 

;ei9 

,887 

1181 

1481 

1759 

2083 

2383 

2707 

3023 

3361 

149 

379 

!631 

•907 

11871483 

1777 

2087 

2389 

,2711  3037 

3371 

151 

383 

!641 

1911 

11931487 

1783 

2089 

2393 

2713 

3041 

3373 

157 

389643 

919 

1201 

1489 

1787 

2099 

2399  2719 

30493389 

163 

397 

|647 

'929 

1213  1493 

1789 

2111 

241112729 

3061 

;3391 

167 

401 

'6531937 

12l7l499!l801 

2113 

2417  2731 

3067|3407 

ART.  162.J  PROPERTIES  OF  NUMBERS.  95 

DIFFERENT   SCALES    OF  NOTATION. 

162.  A  number  expressed  in  the  decimal  notation,  may  be 
changed  to  any  required  scale  of  notation  in  the  following  manner. 
Divide  the  given  number  by  the  radix  of  the  required  scale  con- 
tinually, till  the  quotient  is  less  than  the  radix  ;  then  annex  to  the 
fast  quotient  the  several  remainders  in  a  retrograde  order;  placing 
iphers  where  there  is  no  remainder,  and  the  result  will  be  t/te  num~ 
in  the  scale  required.  (Arts.  43,  44.) 

Ex.  1.  Express  429  in  the  quinary  scale  of  notation. 

Explanation. — By  Dividing  the  given  number          5)429 
by  5,  it  is  evidently  distributed  into   85  parts,          5)  85 — 4 
each  of  which  is  equal  to  5,  with  4  remainder.          5)  17 — 0 
Dividing  again  by  5,  these  parts  are  distributed  3 — 2 

into  17  other  parts,  each  of  which  is  equal  to  5  Ans.  3204 

times  5,  and  the  remainder  is  nothing.  Dividing  by  5  the  third 
time,  the  parts  last  found  are  again  distributed  into  3  other  parts, 
each  of  which  is  equal  to  5  times  5  into  5,  with  2  remainder. 
Thus,  the  given  number  is  resolved  into  3X5X5X5+2X5X5+ 
0X5+4,  or  3204,  which  is  the  answer  required. 

2.  Change  7854  from  the  decimal  to  the  binary  scale. 

Ans.  1111010101110. 

3.  Change  7854  from  the  decimal  to  the  ternary  scale. 

Ans.  101202220. 

4.  Change  7854  from  the  decimal  to  the  quaternary  scale. 

Ans.  1322232. 

5.  Change  7854  from  the  decimal  to  the  quinary  scale. 

Ans.  222404. 

6.  Change  7854  from  the  decimal  to  the  senary  scale. 

Ans.  100210. 

7.  Change  7854  from  the  decimal  to  the  octary  scale. 

Ans.  17256. 

8.  Change  7854  from  the  decimal  to  the  nonary  scale. 

Ans.  11680. 

9.  Change  7854  from  the  decimal  to  the  duodecimal  scale. 

Ans.  4666. 


96  PROPERTIES    OF    NUMBERS.  [SECT.  VI. 

10  Change  35261  from  the  decimal  to  the  quaternary  scale. 

11.  Change  643175  from  the  decimal  to  the  octary  scale. 

12.  Change  175683  from  the  decimal  to  the  septenary  scale. 

13.  Change  534610  from  the  decimal  to  the  octary  scale. 

14.  Change  841568  from  the  decimal  to  the  nonary  scale. 

15.  Change  592835  from  the  decimal  to  the  duodecimal  scale. 

Note. — Since  every  scale  requires  as  many  characters  as  there  are  units  in 
:1  e  radix,  we  will  denote  10  by  t,  and  11  by  e.  Ans.  2470  t  e. 

163*  To  change  a  number  expressed  in  any  given  scale  of 
notation,  to  the  decimal  scale. 

Multiply  the  left  hand  figure  by  the  given  radix,  and  to  the 
product  add  tlie  next  figure  ;  then  multiply  this  sum  by  tlie  radix 
again,  and  to  this  product  add  tlie  next  figure  ;  thus  continue  the 
operation  till  all  tlie  figures  in  t/ie  given  number  Imve  been  employed, 
and  the  last  product  will  be  the  number  in  the  decimal  scale. 

16.  Change  3204  from  the  quinary  to  the  decimal  scale. 

Operation. 

Explanation. — Multiplying  the  left  hand  figure  3204 

by  5,  the  given  radix,  evidently  reduces  it  to  the  5 

next  lower  order ;  for  in  the  quinary  scale,  5  in  17 

an  inferior  order  make  one  in  the  next  superior  5 

order.      For  the   same  reason,  multiplying  this  85 

sum  by  5  again,  reduces   it  to  the  next  lower  5 

order,  <kc.  429  Ans. 

OBS.  This  and  the  preceding  operations  are  the  same  in  principle,  as  reducing 
compound  numbers  from  one  denomination  to  another. 

17.  Change  1522232  from  the  quaternary  to  the  decimal  scale. 

Ans.  7854. 

18.  Change  2546571  from  the  octary  to  the  decimal  scale. 

19.  Change  34120521  from  the  senary  to  the  decimal  scale. 

20.  Change  145620314  from  the  septenary  to  the  decimal  scale. 

21.  Change  834107621  from  the  nonary  to  the  decimal  scale. 

22.  Change  403130021  from  the  quinary  to  the  decimal  scale. 

23.  Change  704400316  from  the  octary  to  the  decimal  scale. 

24.  Change  903124106  from  the  duod ecimal  to  the  decimal  scale. 


ARTS.  163-165.]     PROPERTIES  OF  NUMBERS.  97 

ANALYSIS   OF   COMPOSITE   NUMBERS. 

164.  Every  composite  number,  it  has  been  shown,  may  be 
resolved  into  prime  factors.  (Art.  1C  1.  Prop.  19.) 

Ex.  1.  Resolve  210  into  its  prime  factors. 

Operation.  We  first  divide  the  given  number  by  2,  which 

2)210  is  the  least  number  that  will  divide  it  with" 

3)105  out  a  remainder,  and  which  is  also  a  prime 

6)35"  number.  (Prop.  20.)     We  next  divide  by  3, 

7  then  by  5.     The  several  divisors  and  the  last 

Am.  2,  3,  5,  and  7.  quotient  are  the  prime  factors  required. 

PROOF.  —  2X3X5X7  =  210.     Hence. 

1  G5»  To  resolve  a  composite  number  into  its  prime  factors. 

Divide  the  given  number  by  the  smallest  number  which  will  di- 
vide it  without  a  remainder  ;  then  divide  the  quotient  in  the  same 
way,  and  thus  continue  the  operation  till  a  quotient  is  obtained 
which  can  be  divided  by  no  number  greater  than  1.  The  several 
divisors  with  the  last  quotient,  will  be  thv  prime  factors  required. 
(Art.  161.  Prop.  19.) 

Demonstration.  —  Every  division  of  a  number,  it  is  plain,  resolves  it  into  two 
factors,  viz:  the  divisor  and  dividend.  (Art.  112.)  But  according  to  the  rule, 
the  divisors,  in  every  case,  are  the  smallest  numbers  that  will  divide  the  given 
number  and  the  successive  quotients  without  a  remainder;  consequently  they 
are  all  prime  numbers.  (Art.  101.  Prop.  20.)  And  since  the  division  is  con- 
tinued till  a  quotient  is  obtained,  which  cannot  be  divided  by  any  number 
greater  than  1,  it  follows  that  the  last-  quotient  must  also  be  a  prime  number; 
for,  a  j  rime  number  is  one  which  cannot  be  exactly  divided  by  any  whole 
number  except  a  unit  and  itself.  (Art.  100  Def.  4.) 

OBS.  1.  Since  the  least  divisor  of  every  number  is  a  prime  number,  it  is  evi- 
dent that  a  composite  number  may  be  resolved  into  its  prime  factors,  by  divid- 
ing it  continually  by  any  prime  number  that  will  divide  the  given  number  and 
rhe  quotients  without  a  remainder.  Hence, 

2.  A  composite  number  can  be  divided  by  any  of  its  pr  i  me  factors  without  u 
remainder,  and  by  the  product  of  any  two  or  more  of  them,  but  by  110  ofhcf 
number.  Thus,  the  prime  factors  of  42  are  2,  3,  and  7.  Now  42  can  be  di- 


IBS-  How  do  you  resolve  a  composite  number  into  its  prime  factors  1  Obi.  Will 
the  aaiae  result  be  obtained,  if  we  divide  by  any  of  its  prime  factors  ? 


98  GREATEST    COMMON  [SECT     VI. 

vided  by  2,  3,  and  7;  also  by  2x3,  2X7,  3x7,  and  2x3X^5  but  it  (an  be 
divided  by  no  other  number. 

2.  Resolve  4  and  6  into  their  prime  factouj. 
Solution. — 4  =  2X2  ;  and  6  =  2X3. 

3.  Resolve  8  into  its  prime  factors.     Ans.  8=2X2X2. 

Resolve  the  following   composite   numbers   into   their   prime 
actors : 

4.  9.  22.  34.  40.  57.  58.  81. 

5.  10.  23.  35.  41.  58.  59.  82. 

6.  12.  24.  36.  42.  60.  60.  84. 

7.  14.  25.  38.  43.  62.  61.  85. 

8.  15.  26.  39.  44.  63.  62.  86. 

9.  16.  27.  40.  45.  64.  63.  87. 

10.  18.  28.  42.  46.  65.  64.  88. 

11.  20.  29.  44.  47.  66.  65.  90. 

12.  21.  30.  45.  48.  68.  66.  91. 

13.  22.  31.  46.  49.  69.  67.  92. 

14.  24.  32.  48.  50.  70.  68.  93. 

15.  25.  33.  49.  51.  72.  69.  94. 

16.  26.  34.  50.  52.  74.  70.  95. 

17.  27.  35.  51.  53.  75.  71.  96. 

18.  28.  36.  52.  54.  76.  72.  98. 

19.  30.  37.  54.  55.  77.  73.  99. 

20.  32.  38.  55.  56.  78.  74.  100. 

21.  33.  39.  56.  57.  80.  75.  108. 

76.  Resolve  120  and  144  into  their  prime  factors. 

77.  Resolve  180  and  420  into  their  prime  factors. 

78.  Resolve  714  and  836  into  their  prime  factors. 

79.  Resolve  574  and  2898  into  their  prime  factors. 
BO.  Resolve  11492  and  180  into  their  prime  factors. 

81.  What  are  the  prime  factors  of  650  and  1728  ? 

82.  What  are  the  prime  factors  of  1492  and  8032  ? 

83.  What  are  the  prime  factors  of  4604  and  16806  ? 

84.  What  are  the  prime  factors  of  71640  and  20324  ? 

85.  What  are  the  prime  factors  of  84705  and  65948  ? 

86.  What  are  the  prime  factors  of  92353  and  8T3784  ? 


ARTS.  166-168.]  DIVISOR.  99 

GREATEST  COMMON  DIVISOR. 

166»  A  common  divisor  of  two  or  more  numbers,  is  a  num- 
ber which'  will  divide  each  of  them  without  a  remainder.  Thus 

2  is  a  common  divisor  of  6,  8,  12,  16,  18,  &c. 

167.  The  greatest  common  divisor  of  two  or  more  numbers, 
is  the  greatest  number  which  will  divide  th  em  without  a  remainder. 
Thus  6  is  the  greatest  common  divisor  of  12,  18,  24,  and  30. 

OBS.  A  common  divisor  is  sometimes  called  a  common  measure.  It  will  be 
eefcn  that  a  common  divisor  of  two  or  more  numbers,  is  simply  a  factor  which 
is  common  to  those  numbers,  and  the  greatest  common  divisor  is  the  greatest 
factor  common  to  them.  Hence, 

1G8»  To  find  a  common  divisor  of  two  or  more  numbers. 

Resolve  each  number  into  two  or  more  factors,  one  of  which  sJiall 
be  common  to  all  the  given  numbers. 

Or,  resolve  the  given  numbers  into  their  prime  factors,  then  if 
the  same  factor  is  found  in  each,  it  will  be  a  common  divisor.  (Art. 
1C5.  Obs.  2.) 

OBS.  If  the  given  numbers  have  not  a  common  factor ,  they  cannot  have  a 
common  divisor  greater  than  a  unit ;  consequently  they  are  either  prime  num- 
bers, or  are  prime  to  each  other.  (Art.  1150.  Def.  3.  Obs.  2.) 

Note. — The  following  facts  may  assist  the  learner  in  finding  common  di- 
visors : 

1.  Any  numbe*  ending  in  0,  or  an  even  number,  as  2,  4,  6,  &c.,  may  be 
divided  by  2. 

2.  Any  number  ending  in  5  or  0,  may  be  divided  by  5. 

3.  Any  number  ending  in  0,  may  be  divided  by  10. 

4.  When  the  two  right  hand  figures  are  divisible  by  4,  the  whole  number 
may  be  divided  by  4. 

5.  If  the  three  right  hand  figures  of  any  number  are  divisible  by  8,  the 
whole  is  divisible  by  8. 

Ex.  1.  Find  a  common  divisor  of  6,  15,  and  21. 

Solution. — 6  =  3X2;  15  =  3X5;  and  21  =  3x7.     The  factor 

3  is  common  to  each  of  the  given  numbers,  and  is  therefore  a 
common  divisor  of  them. 

<iuKST. — 166.  What  is  a  common  divisor  of  two  or  more  numbers  ?  167,  What  is  the 
greatest  common  divisor  of  two  or  more  numbers  ?  Obs.  What  is  a  common  divisor  some- 
times called!  168.  How  do  you  find  a  common  divisor  of  two  or  more  numbers]  Obi.  I" 
two  given  numbers  have  not  a  common  factor,  what  is  true  as  to  a  common  divisor  ? 

5* 


tOO  GREATEST  COMMON          [SECT.  VI. 

2.  Find  a  common  divisor  of  15,  18,  24,  and  36. 

3.  Find  a  common  divisor  of  14,  28,  42,  and  35. 

4.  Find  a  common  divisor  of  10,  35,  50,  75,  and  30. 
6.  Find  a  common  divisor  of  82,  118,  and  146. 

6.  Find  a  common  divisor  of  42  and  66.     Ans.  2,  3,  or  6. 

169.  It  will  be  seen  from  the  last  example  that  two  numbers 
may  have  more  than  one  common  divisor.  In  many  cases  it  is 
highly  important  to  find  the  greatest  divisor  that  will  divide  two 
or  more  given  numbers  without  a  remainder. 

7.  What  is  the  greatest  common  divisor  of  35  and  50  ? 

Operation.  Dividing  50   by  35,  the  remainder    is  15, 

35)50(1  then  dividing  35  (the  preceding  divisor)  by 

35  15  (the  last  remainder)  the  remainder  is  5  ; 

15)35(2  finally,  dividing  15  (the  preceding  divisor)  by 

30  5  (the  last  remainder)  nothing  remains ;  con- 

5)15(3  sequently  5,  the  last  divisor,  is  the  greatest 

15  common  divisor.     Hence, 

1 7  O.  To  find  the  greatest  common  divisor  of  two  numbers. 

Divide  the  greater  number  by  the  less  ;  then  divide  the  preceding 
divisor  by  the  last  remainder,  and  so  on,  till  nothing  remains. 
The  last  divisor  will  be  the  greatest  common  divisor. 

When  there  are  more  than  two  numbers  given. 

first  find  the  greatest  common  divisor  of  any  two  of  tliem ; 
then,  that  of  the  common  divisor  tJms  obtained  and  of  another 
given  number,  and  so  on  through  all  the  given  numbers.  The  last 
common  divisor  found,  will  be  the  one  required. 

Demonstration. — Since  5  is  a  measure  of  the  last  dividend  15,  in  the  preced- 
ing solution,  it  must  therefore  be  a  measure  of  the  preceding  dividend  35;  be- 
cause 35— 2xl5-(-5;  and  35  is  oae  of  the  given  numbers.  Now,  since  5 
measures  15  and  35,  it  must  also  measure  their  sum,  viz :  35-J-15,  or  50,  which 
18  the  other  given  number.  (Art.  161.  Prop.  13.)  In  a  similar  manner  it  may 
be  shown  that  the  last  divisor  will,  in  all  cases,  be  the  greatest  common  divisor. 

Note. — Numbers  which  have  no  common  measure  greater  than  1,  are  said  to. 
be  incommensuro3)le.  Thus  17  and  29  are  incommensurable. 


.— 170.  How  find  the  greatest  common  divisor  of  two  numbers  1  Of  more  than  twol 


ARTS.  169-171.]  DIVISOR.         .      •*.  101 

8.  What  is  the  greatest  common  divisor  of  285  and  465  ? 

9.  What  is  the  greatest  common  divisor  of  532  and  1274  ? 

10.  What  is  the  greatest  common  divisor  of  888  and  2775  ? 

11.  What  is  the  greatest  common  divisor  of  2145  and  3471  ? 

12.  What  is  the  greatest  common  divisor  of  1879  and  2425  ? 

13.  What  is  the  greatest  common  divisor  of  75,  125,  and  100  ? 

Suggestion. — Find  the  greatest  common  divisor  of  75  and  125, 
which  is  25.  Then  that  of  25  and  160.  Ans.  5. 

14.  What  is  the  greatest  common  divisor  of  183,  3996,  108  ? 

15.  What  is  the  greatest  common  divisor  of  672, 1440,  and  3472  ? 

16.  What  is  the  greatest  common  divisor  of  30,  42,  and  66  ? 

Analysis. — By  resolving  the  given  num-  Operation* 

bers  into  their  prime  factors,  (Art.  165,)         30  =  2X3X5 
we  find  that  the  factors  2  and  3  are  both         42  =  2X3x7 
common  divisors  of  them.     But  we  have         66  =  2X3X11 
seen   that    a   composite    number   can  be     Now  2X3  =  6  Ans. 
divided    by  the    product   of  any   two   or 

more  of  its  prime  factors  ;  (Art.  165.  Obs.  2  ;)  consequently  30, 
42,  and  66  can  all  be  divided  by  2X-3  ;  for  2X3  is  the  product 
of  two  prime  factors  common  to  each.  And  since  they  are  the 
onl\r  factors  common  to  the  given  numbers,  their  product  must 
be  the  greatest  common  divisor  of  them.  Hence,  we  deduce  a 

171*  Second  Method  of  finding  the  greatest  common  divisor 
of  two  or  more  numbers. 

Resolve  the  given  numbers  into  their  prime  factors,  and  the  con- 
tinued product  of  those  factors  which  are  common  to  each,  will  be 
the  greatest  common  divisor. 

OBS.  If  the  given  numbers  have  but  one  common  factor,  that  factor  itself  ia 
the  greatest  common  divisor. 

17.  What  is  the  greatest  common  divisor  of  105  and  165  ? 

18.  What  is  the  greatest  common  divisor  of  36,  60,  and  108 

19.  What  is  the  greatest  common  divisor  of  108,  126,  and  162  ? 

20.  What  is  the  greatest  common  divisor  of  140,  210,  and  315  ? 

21.  What  is  the  greatest  common  divisor  of  24,  i2,  54,  and  60  ? 

22.  What  is  the  greatest  common  divisor  of  5  C,  84, 1 40,  and  168  ? 


102  LEAST    COMMON  [SECT.  VI. 

LEAST   COMMON   MULTIPLE. 

172.  One  number  is  said  to  be  a  multiple  of  another,  when 
the  former  can  be  divided  by  the  latter  without  a  remainder 
(Art.  160.  Def.  8.)     Hence, 

173.  A  common  multiple  of  two  or  more  numbers,  is  a  num- 
ber which  can  be  divided  by  eacl  of  them  withe  at  a  remainder. 
Thus,  12  is  a  common  multiple  of  2,  3,  and  4;  15  is  a  common 
multiple  of  3  and  5,  (fee. 

OBS.  A  comvwn  multiple  is  always  a  composite  number,  of  which  each  of 
the  given  numbers  must  be  t  "actor ;  otherwise  it  could  not  be  divided  by 
them.  (Art.  1G5.  Obs.  2.) 

174.  The  continued  product  of  two  or  more  given  numbers 
will  always  form  a  common  multiple  of  those  numbers.     The  same 
numbers  may  have  an  unlimited  number  of  common  multiples ; 
for,  multiplying  their  continued  product  by  any  number,  will  form 
a  new  common  multiple,  (Art.  161.  Prop.  14.) 

175.  The  least  common  multiple  of  two  or  more  numbers,  is 
the  least  number  which  can  be  divided  by  each  of  them  without  a 
remainder.     Thus,  12  is  the  least  common  multiple  of  4  and  6,  for 
it  is  the  least  number  which  can  be  exactly  divided  by  them. 

OBS.  The  least  common  multiple  of  two  OT  more  numbers,  is  evidently 
composed  of  all  the  prime  factors  of  each  of  the  given  numbers  repeated  once, 
and  only  once.  For,  if  it  did  not  contain  all  the  prime  factors  of  any  one  of 
the  given  numbers,  it  could  not  be  divided  by  tbat  number.  (Art.  165.  Obs.  2.) 
On  the  other  hand,  if  any  prime  factor  is  employed  more  limes  than  it  is  re- 
peated as  a  factor  in  some  one  of  the  given  numbers,  then  it  woild  not  be  th» 
least  common  multiple. 

Ex.  1.  What  is  the  least  common  multiple  of  10  and  15  ? 

Analysis. — 10  =  2X5,  and  15  =  3X5.  The  prime  factors  of 
the  given  numbers  are  2,  5,  3,  and  5.  Now  since  the  .'actor  5 
>ccurs  once  in  each  number,  we  may  therefore  cancel  L  in  one 


. — 172.  When  is  one  number  said  to  be  a  multiple  of  anothejf?  173  What  *s  a 
common  multiple?  174.  How  may  a  common  multiple  of  two  t>r  more  r  »mbers»  t« 
formed  ?  How  many  common  multiples  may  there  be  of  at  y  giver  lumbers  ?  176.  What 
if  the  least  common  multiple  of  two  or  more  nuir  oers  1 


ARTS.  1T2--176.J  MULTIPLE.  103 

instance,  and  the  continued  product  of  the  remaining  factors  2X3 
X5,  or  30,  will  be  the  least  common  multiple. 

Operation.  We  first  divide  both  the  numbers  by  5 

5)10  "  15  in  order  to  resolve  them  into  prime  fac- 

2  "     3  tors.  (Art.  175.  Obs.)     Thus,  all  the  dif- 

5X2X3=30^5.  ferent  factors  of  which  the  given  num- 
bers are  composed,  are  found  in  the  divisor  and  quotients  onct, 
and  only  once.  Therefore  the  product  of  the  divisor  and  quotients 
5X2X3,  is  the  least  common  multiple  required.  Hence, 

176.  To  find  the  least  common  multiple  of  two  or  more 
numbers. 

Write  the  given  numbers  in  a  line  with  two  points  between  them. 
Divide  by  the  smallest  number  which  will  divide  any  two  or  more 
of  them  without  a  remainder,  and  set  the  quotients  and  the  undivided 
numbers  in  a  line  beloiv.  Divide  this  line  and  set  down  the  re- 
sults as  before  j  thus  continue  the  operation  till  there  are  no  two 
numbers  which  can  be  divided  by  any  number  greater  than  1.  The 
continued  product  of  the  divisors  into  the  numbers  in  the  last  line, 
will  be  the  least  common  multiple  required. 

OBS.  1.  We  have  seen  that  the  least  dimsor  of  every  number  isaprimt  num- 
ber ;  hence,  dividing  by  the  smallest  number  which  will  divide  two  or  more  of 
the  given  numbers,  is  dividing  them  by  a  prime  number.  (Art.  161.  Prop. '20.) 

The  result  will  evidently  be  the  same,  if,  instead  of  dividing  by  the  smallest 
number,  we  divide  the  given  numbers  by  any  prime  number,  that  will  divide 
two  or  more  of  them,  without  a  remainder. 

2.  The  preceding  operation,  it  will  be  seen,  resolves  the  given  numbers  into 
their  prime  factors,  (Art.  1G5,)  then  multiplies  all  the  different  factors  together, 
taking  each  factor  as  many  times  in  the  product,  as  are  equal  to  the  greatest 
number  of  times  it  is  found  in  either  of  the  given  numbers. 

3.  If  the  given  numbers  are  prime  numbers,  or  are  prime  to  each  other,  the 
continued  product  of  the  numbers  themselves  will  be  their  least  common  mul- 
tiple. (Art.  168.  Obs.)     Thus,  the  least  common  multiple  of  5  and  7  is  35;  of 
8  and  9  is  72. 

UUKST. — 176.  How  is  the  least  common  multiple  of  two  or  more  numbers  found t 
Obs.  If  the  Riven«numbers  are  prime,  or  are  prime  to  each  other,  what  is  the  least  com 
mon  multiple  of  them  1  176.  a.  Upon  what  principle  does  this  rule  depend  1  Obs.  Why 
do  you  divide  by  the  smallest  number  that  will  divide  two  or  more  of  the  given  numben 
without  a  remainder  1 


104  LEAST    COMMON  [SECT.  VI 

Ex.  2,  What  is  the  least  common  multiple  of  6,  8,  and  12  ? 

Analysis. — By  resolving  the  given  numbers  Operation. 

into  their  prime  factors,  it  will  be  seen  that  2       6  =  2X;3 
is  found  once  as  a  factor  in  6  ;  twice  in  1 2  ;  and       8  —  2x2X2 
three  times  in  8.     It  must  therefore  be  taken     12  =  2X2X3 
three  times'in.  the  product.     Again,  3  is  a  lac-     2X2X2X3=24 
tor  of  6,  and  12,  consequently  it  must  be  taken  only  once  in  the 
product.  (Art.  176.  Obs.  2.)     Thus,  2X2X2X3  =  24  Ans. 

Ex.  3.  What  is  the  least  common  multiple  of  12,  18.  and  36  ? 

First  Operation.  Second  Operation.  Third  Operation. 

2)12  "  18  "  36  9)12  "  18   *'  36  12)12  "   18  "  36 

2)  6  "     9  ""18  2)12  "     2  "~4  3)   1   "   18  "     3 
3)~3~ 9~" 9  2)  6  "      1   "     2  1   " 6~" 1 

3)  1   "     3~"     3  3  " r~"     1       And  12X3X6=216. 
1   '•      F~" T         Now  9X2X2X3=108. 

2X2X3X3  =  36  Ans. 

Explanation. — In  the  first  operation,  we  divide  by  the  smallest 
numbers  which  will  divide  any  two  or  more  of  the  given  numbers 
without  a  remainder,  and  the  product  of  the  divisors,  &c.,  is  36, 
which  is  the  answer  required. 

In  the  second  and  third  operations,  we  divide  by  numbers  that 
will  divide  two  or  more  of  the  given  numbers  without  a  remainder, 
and  in  both  cases,  obtain  erroneous  answers. 

Note. — It  will  be  seen  from  the  second  and  third  operations  above,  that 
tl  dividing  by  any  number,  which  will  divide  two  or  more  of  the  given  num- 
bers without  a  remainder,"  according  to  the  rule  given  by  some  authors,  Joes 
not  always  give  the  least  common  multiple  of  the  numbers. 

176.  a.  The  reason  of  the  preceding  rule  depends  upon  the 
principle  that  the  least  common  multiple  of  any  two  or  more  num- 
bers, is  composed  of  all  the  prime  factors  of  the  given  numbeis, 
each  taken  as  many  times,  as  are  equal  to  the  greatest  numle~  of 
times  it  is  found  in  either  of  the  given  numbers.  (Art.  175.  Obs.) 

Note. — 1.  The  reason  for  dividing  by  the  smallest  number,  is  because  tha 
divisor  may  otherwise  be  a  composite  number,  (Art.  1G1.  Prop.  20.)  and  have 
a  factor  common  to  some  one  of  the  quotients,  or  undivided  numbers  in  the 
last  line;  consequently  the  continued  product  of  them  wotJd  be  too  large  for 


ARTS.  176.  #.  177.]  MULTIPLE.  105 

the  kast,  common  multiple.  (Art.  175.  Obs.)  Thus,  in  t  e  second  op:/ration  the 
divisor  9,  is  a  composite  number,  containing  the  factor  3  common  to  the  3  in 
the  quotient;  consequently  the  product  is  three  times  too  lartfc.  In  the  third 
operation  the  divisor  12,  is  a  composite  number,  and  contains  the  factor  6  com- 
mon to  the  6  in  the  quotient;  therefore  the  product  is  six  times  too  large. 

2.  The  object  of  arranging  the  given  numbers  in  a  line,  is  that  all  of  them 
may  be  resolved  into  their  prime  factors  at  the  same  time  ;  and  also  to  present 
at  a  giance  the  factors  which  compose  the  least  common  multiple  required. 

4.  Find  the  least  common  multiple  of  6,  9,  and  15. 

5.  Find  the  least  common  multiple  of  8,  16,  18,  and  24. 
G.  Find  the  least  common  multiple  of  9,  15,  12,  6,  and  5. 

7.  Find  the  least  common  multiple  of  5,  10,  8,  18,  and  15, 

8.  Find  the  least  common  multiple  of  24,  16,  18,  and  20. 

9.  Find  the  least  common  multiple  of  36,  25,  60,  72,  and  36. 

10.  Find  the  least  common  multiple  of  42,  12,  84,  and  72. 

11.  Find  the  least  common  multiple  of  27,  54,  81,  14,  and  63 

12.  Find  the  least  common  multiple  of  7,  11,  13,  3,  and  5. 

177.  The  process  of  finding  the  least  common  multiple 
may  often  be  shortened,  by  canceling  every  number  which  will 
divide  any  other  given  number,  without  a  remainder,  and  also 
those  which  will  divide  any  other  number  in  the  same  line.  The 
least  common  multiple  of  the  numbers  that  remain,  will  be  the  an- 
swer required. 

OBS.  By  attention  and  practice,  the  student  will  be  able  to  discover,  by  in- 
spection, the  least  common  multiple  of  numbers,  when  they  are  not  large. 

13.  Find  the  least  common  multiple  of  4,  6,  10,  8,  12,  and  15. 

Operation.  Since  4  and  6,  will  exactly  di- 

2)^  "  0  "  10  "  8  "  12  "  15  vide  8,  and  12,  we  cancel  them. 


2)       0  "  4  "    6  "  15  Again,  since  5  in  the  second  line 

3)    2  "    3  "  15  will  exactly  divide  1  5  in  the  same 

2  "     1  "    5  line,  we  therefore  cancel  it,  and 

Now,  2X2X3  X2  X  5  =  120  Ans.  proceed  with  the  remaining  num- 

bers as  before. 

14.  Find  the  least  common  multiple  of  9,  12,  72,  36,  and  144. 

15.  Find  the  least  common  multiple  of  8,  12   20,  24,  and  25. 

16.  Find  the  least  common  multiple  of  1,  2,  3,  4,  5,  6,  7,  8,  9, 


106  COMMON  [SECT.  VI. 

17.  Find  the  least  common  multiple  of  33,  12,  84,  and  7. 

18.  Find  the  least  common  multiple  of  54,  81,  63,  and  14. 

19.  Find  the  least  common  multiple  of  72, 120, 180,  24,  and  36. 

177.  a.  The  least  common  multiple  of  two  or  more  numbers 
may  also  be  found  in  the  following  manner. 

First  find  the  greatest  common  divisor  of  two  of  the  given  num- 
bers ;  by  this  divide  one  of  these  two  numbers,  and  multiply  the 
quotient  by  the  other.  Then  perform  a  similar  operation  on  the 
product  and  another  of  the  given  numbers ;  thus  continue  the  pro- 
cess until  all  of  the  given  numbers  have  been  employed,  and  tJie 
final  result  will  be  the  least  common  multiple  required. 

20.  What  is  the  least  common  multiple  of  24,  16,  and  12  ? 

Solution. — By  inspection,  we  find  the  greatest  common  divisor 
of  24  and  16,  is  8.     Now  24-7-8  =  3;  and  3X16  =  48.     Again,' 
the  greatest  common  divisor  of  48  and  12,  is  12.     Now  48-f-12 
=4;  and  4X12=48.  Ans. 

PROOF. — Resolving  the  given  numbers  into  their  prime  factors, 
24  =  2X2X2X3;  16  =  2X2X2X2;  and  12  =  2X2X3;  (Art. 
165  ;)  consequently,  2X2X2x2X3=48,  the  least  common  mul- 
tiple. (Art.  175.  Obs.) 

OBS.  The  reason  of  this  rule  depends  upon  the  principle,  that  if  the  product 
of  any  two  numbers  be  divided  by  any  factor  which  is  common  to  both,  the 
quotient  will  be  a  common  multiple  of  the  two  numbers.  Thus,  if  48,  the 
product  of  G  and  8,  be  divided  by  2,  a  factor  of  both,  the  quotient  24,  will  be 
a  multiple  of  each,  since  it  may  be  regarded  either  as  8  multiplied  by  the  quo- 
tient of  6  by  the  factor  2,  or  as  6  multiplied  by  the  quotient  of  8  by  the  same 
factor.  Hence,  it  is  obvious,  that  the  greater  the  common  measure  is,  the  less 
will  be  the  multiple ;  and,  consequently,  the  greatest  common  measure  will 

oduce  the  least  common  multiple. 

When  the  common  multiple  of  the  first  two  numbers  is  found,  it  is  evident, 

at  any  number  which  is  a  common  multiple  of  it  and  the:  third  number,  will 
?*»  a  multiple  of  the  first,  second,  and  third  numbers. 

21.  What  is  the  least  common  multiple  of  75,  120,  and  300  ? 

22.  What  is  the  least  common  multiple  of  96,  144,  and  720  ? 

23.  What  is  the  least  common  multiple  of  256,  512,  and  1728  ? 

24.  What  is  the  least  common  multiple  of  S'/5,  85  J,  and  3400  ? 


ARTS.    77.  a- 181. ]         FRACTIONS.  107 

SECTION    VII. 
FRACTIONS. 

ART.  178.  When  a  number  or  thing  is  divided  into  two  equal 

parts,  one  of  those  parts  is  called  one  half.     If  the  number  or 

hing  is  divided  into  three  equal  parts,  one  of  the  parts  is  called 

one  third  ;  if  it  is  divided  into  four  equal  parts,  one  of  the  parts 

is  called  one  fourth,  or  one  quarter  ;  and,  universally, 

When  a  number  or  thing  is  divided  into  equal  parts,  the  parts 
take  tJwir  name  from  the  number  of  parts  into  which  the  thing  or 
number  is  divided. 

1 7  9»  The  value  of  one  of  these  equal  parts  manifestly  depends 
upon  the  number  of  parts  into  which  the  given  number  or  thing 
is  divided.  Thus,  if  an  orange  is  successively  divided  into  2,  3, 
4,  " ,  6,  &c.,  equal  parts,  the  thirds  will  be  less  than  the  halves ; 
the  fourths,  than  the  thirds ;  the  fifths,  than  the  fourths,  &c. 

OBS.  A  half  of  any  number  is  equal  to  as  many  units,  as  2  is  contained 
times  in  that  number ;  a  third  of  a  number  is  equal  to  as  many,  as  3  is  con- 
tained times  in  the  given  number ;  a  fourth  is  equal  to  as  many,  as  4  is  con- 
tained in  the  number,  &c. 

1  8O«  When  a  number  or  thing  is  divided  into  tqual  parts, 
these  parts  are  called  FRACTIONS. 

OBS.  Fractions  are  used  to  express  parts  of  a  collection  of  things,  as  well  as 
of  a  single  thing ;  or  parts  of  any  number  of  units,  as  well  as  of  one  unit. 
Thus,  we  speak  of  •%  of  six  oranges ;  &  of  75;  &Ct  in  this  case  the  collection, 
or  number  to  be  divided  into  equal  parts,  is  regarded  as  a  whole. 

181.  Fractions  are  divided  into  two  classes,  Common  and 
Decimal.  For  the  illustration  of  Decimal  Fractions,  see  Sec- 
tion IX. 

QUEST.— 178.  What  is  meant  by  one  half1?  What  is  meant  by  one  third?  What  is 
meant  by  a  fourth  ?  What  is  meant  by  fifths  ?  By  sixths  ?  How  many  sevenths  make 
a  whole  one  ?  How  many  tenths  ?  What  is  meant  by  twentieths  ?  By  hundreds  ?  When 
a  number  or  thing  is  divided  into  equal  parts,  from  what  do  the  parts  take  their  name  1 
179.  Upon  what  does  the  value  of  one  of  these  equal  parts  depend  7  180.  What  are  frao 
tkras  ?  181.  Into  how  many  classes  are  fractions  divided  1 


108  FRACTIONS.  [SECT.  VII. 


!2.  Common  Fractions  are  expressed  by  two  numbers,  one 
plat  ^d  over  the  other,  with  a  line  between  them.  One  half  is 
written  thus  •£  ;  one  third,  ^  ;  one  fourth,  •}•  ;  nine  tenths,  -ft-  ; 
thirteen  forty-fifths,  •$•£,  &c. 

The  number  below  the  line  is  called  the  denominator,  and  shows 
into  Jtow  tnany  parts  the  number  or  thing  is  divided. 

The  number  above  the  line  is  called  the  numerator,  and  shows 
Itow  many  parts  are  expressed  by  the  fraction.     Thus,  in  the  frac 
tion  -f,  the  denominator  3,  shows  that  the  number  is  divided  into 
three  equal  parts  ;  the  numerator  2,  shows  that  two  of  those  parts 
are  expressed  by  the  fraction. 

The  denominator  and  numerator  together  are  called  the  terms 
of  the  fraction. 

OBS.  1.  The  term  fraction,  is  of  Latin  origin,  and  signifies  broken,  or  sepa 
rated  into  parts.  Hence,  fractions  are  sometimes  called  broken  numltcrs, 

2.  Common  fractions  are  often  called  vulgar  fractions.     This  term,  however, 
is  very  properly  falling  into  disuse. 

3.  The  number  below  the  line  is  called  the  denominator,  because  it  gives  the 
name  or  denomination  to  the  fraction  ;  as,  halves,  thirds,  fifths,  &c. 

The  number  above  the  line  is  called  the  numerator,  because  it  numbers  the 
parts,  or  shows  how  many  parts  are  expressed  by  the  fraction. 

183*  A  proper  fraction  is  a  fraction  whose  numerator  is  less 
than  its  denominator  ;  as,  •£,  -f  ,  -f-. 

An  improper  fraction  is  one  whose  numerator  is  equal  to,  or 
greater  than  its  denominator  ;  as,  -f,  f. 

A  mixed  number  is  a  whole  number  and  a  fraction  expressed 
together  ;  as,  4-f,  25-Ji. 

A  simple  fraction  is  a  fraction  which  has  but  one  numerator  and 
one  denominator,  and  may  be  proper,  or  improper  ;  as,  f  ,  f  . 

A  compound  fraction  is  a  fraction  of  a  fraction  ;  as,  -f  of  -f-  of  f> 


QUEST.  —  182.  How  are  common  fractions  expressed  ?  What  is  the  number  below  the 
line  called?  What  does  it  show?  What  is  the  number  above  the  line  called?  What 
does-  it  show  ?  What  are  the  denominator  and  numerator,  taken  together,  called  ? 
Obs.  What  is  the  meaning  of  the  term  fraction  ?  What  are  common  fractions  sometimes 
called?  Why  is  the  lower  number  called  the  denominator1  Why  is  the  upper  one 
called  the  numerator  ?  183.  What  is  a  proper  fraction  7  An  improper  fraction  1  A  Mixed 
number  1  A  simple  fraction  ?  A  compound  fraction  1 


ARTS.  182-188.]  FRACTIONS.  109 

A  complex  fraction  is  one  which  has  a  fraction  in  its  numerator 

2^    4      2-1 

or  denominator,  or  in  both  ;  as,   — »  — ,  —  • 

5      5-3    84- 

1 8  4-.  Fractions,  it  will  be  seen  both  from  the  definition  and 
the  mode  of  expressing  them,  arise  from  division,  and  may  be 
treated  as  exp:essions  of  unexecuted  division.  The  numerator  an- 
swers to  the  dividend,  and  the  denominator  to  the  divisor.  (Arts, 
25,  182.)  Hence, 

1  85»  The  value  of  a  fraction  is  the  quotient  of  the  numerator 
divided  by  the  denominator.  Tims,  the  value  of  f  is  two  ;  of  \  is 
one  ;  of  \  is  one  third,  &c.  Hence, 

186»  If  the  denominator  remains  the  same,  multiplying  the 
numerator  by  any  number,  multiplies  the  value  of-  the  fraction  bi 
that  number.  For,  since  the  numerator  and  denominator  answe 
to  the  dividend  and  divisor,  multiplying  the  numerator  is  the  sama 
as  multiplying  the  dividend.  But  multiplying  the  dividend,  we 
have  seen,  multiplies  the  quotient,  (Art.  141,)  which  is  the  same 
as  the  value  of  the  fraction.  (Art.  185.)  Thus,  the  value  of  f=2  ; 
now,  multiplying  the  numerator  by  3,  the  fraction  becomes  -^ 
whose  value  is  6,  and  is  the  same  as  2X3. 

187.  Dividing  the  numerator  by  any  number,  divides  the  value 
of  the  fraction  by  that  number.     For,  dividing  the  dividend,  divides 
the  quotient.  (Art.  142.)     Thus,  -J=2  ;  now  dividing  the  numera- 
tor by  2,  the  fraction  becomes  -f,  whose  value  is  1,  and  is  the  same 
as  2-^2.     Hence, 

OBS.  With  a  given  denominator,  the  greater  the  numerator,  the  greater  will 
/     oe  the  value  of  the  fraction. 

188.  If  the  numerator  remains  the  same,  multiplying  the  de- 
nominator by  any  number,  divides  the  value  of  the  fraction  by  that 
number.     For,  multiplying  the  divisor,  we  have  seen,  divides  th 

QTTKST. — What  is  a  complex  fraction  ?  184.  From  what  tin  fractions  arise  ?  185.  "That 
is  the  value  of  a  fraction  ?  186.  What  is  the  effect  of  multiplying  the  nnmeritnr.  <vhile 
the  denominator  remains  the  same  ?  Explain  the  reason.  187.  Wh.  /  is  the  effect  of  di 
viding  the  numerator?  Obs.  With  a  given  denominator,  what  is  the  effect  ol  increasing 
*he  numerator  7  188.  What  is  the  effect  of  multiplying  the  denominator  1 


110  FRACTIONS.  [SECT.    VII. 

quotient.  (A.  1.143.)  Thus,  -^=4;  now  multiplying  the  denom- 
inator by  2,  the  fraction  becomes  -f£,  whose  value  is  2,  and  is  the 
same  as  4-^-2. 

189,  Dividing  the  denominator  by  any  number,  multiplies  the 
value  of  the  fraction  by  that  number.     For,  dividing  the  divisor 
multiplies  the  quotient.  (Art.  144.)     Thus,  *£-=4 ;  now  dividing 
the  denominator  by  2,  the  fraction  becomes  -^S  whose  value  is  3, 

nd  is  the  same  as  4X2.     Hence, 

OBS.  With  a  given  numerator,  the  greater  the  denominator,  the  less  wiii  oe 
the  value  of  the  fraction. 

1 9O.  It  is  evident  from  the  preceding  articles,  that  multiply- 
ing the  numerator  by  any  number,  has  the  same  effect  on  the  value 
of  the  fraction,    as   dividing   the  denominator  by  that   number. 
(Arts.  186,  189.)     And, 

Dividing  the  numerator  has  the  same  effect,  as  multiplying  the 
denominator.  (Arts.  187,  188.) 

OBS.  It  will  be  observed,  that  multiplying  or  dividing  the  numerator  of  a 
fiaction,  has  the  same  effect  upon  its  value,  as  the  same  operation  has  upon 
a  whole  number ;  but,  the  effect  of  multiplying  or  dividing  the  denominator  is 
exactly  contrary  to  that  of  the  same  operation  upon  a  whole  number. 

191,  If  the  numerator  and  denominator  are  both  multiplied 
or  both  divided  by  the  same  number,  the  value  of  the  fraction  will 
not  be  altered.  (Art.  146.)     Thus,  -^=3;  now  if  the  numerator 
and  denominator  are  both  multiplied  by  2,  the  fraction  becomes 
^gS  whose  value  is  3.     If  both  terms  are  divided  by  2,  the  frac- 
tion becomes  f,  whose  value  is  3  ;  that  is,  -L4-a=-2^-=f =3. 

192.  Since  the  value  of  a  fraction  is  the  quotient  of  the 
numerator  divided  by  the  denominator,  it  follows, 

If  the  numerator  and  denominator  are  equal,  the  value  is  a  unit 
or  one.  Thus,  •§•=!,  -f=l,  &c. 

QUEST. — 189.  W.iat  is  the  effect  of  dividing  the  denomina  OT  ?  Why?  Obs,  With  a 
given  numerator,  what  is  the  effect  of  increasing  the  denominator  1  190.  What  may  be 
done  to  the  denominator  to  produce  the  same  effect  on  the  value  of  the  fraction,  as  mal 
tiplying  the  numerator  by  any  given  number?  What,  to  product  the  same  effect  as  divid- 
ing the  numerator  by  any  given  number  ?  191.  What  is  the  effect  if  the  numerator  and 
denominator  are  both  multiplied,  or  both  divided  by  the  same  number?  192  When  tho 
numerator  and  denominator  are  equal,  what  is  the  value  of  the  fraction? 


ARTS.  189-194.]  nt  ACTIONS.  Ill 

If  the  numerator  is  greater  than  the  denominator,  the  value  is 
greater  than  one.  Thus,  £=2,  £=l-f. 

If  the  numerator  is  leas  than  the  denominator,  the  value  is  less 
than  one.  Thus,  £=1  third  of  1,  £=4  fifths  of  1. 

193*  FJ  actions  may  be  added,  subtracted,  multiplied,  and 
divided,  as  well  as  whole  numbers.  But,  in  order  to  perform 
the  se  operations,  it  is  often  necessary  to  make  certain  changes  in 
llie  terms  of  the  fractions. 

OBS  It  is  evident  that  any  changes  may  be  made  in  the  terms  of  a  fraction, 
which  do  not  alter  the  quotient  of  the  numerator  divided  by  the  denominator; 
for,  it'  the  quotient  is  not  altered,  the  value  remains  the  same.  Thus.  the.  terms 
of  the  fraction  •£•  may  be  changed  into  -^,  -f-,  -L/-,  &c.,  without  altering  its  value  j 
for  in  each  case  the  quotient  of  the  numerator  divided  by  the  denominator  is  2. 
Hence,  for  any  given  fraction,  we  may  substitute  any  other  fraction,  which 
wih  give  the  same  quotient. 


REDUCTION   OF    FRACTIONS. 

194.  The  process  of  changing  the  terms  of  a  fraction  into 
others,  without  altering  its  value,  is  called  REDUCTION  OF  FRAC- 
TIONS. 

CASE    I. 

Bx.  1.  Reduce  £#  to  its  lowest  terms. 

First  Operation.  Dividing  both  terms  of  the 

2K-u=-i6o  :  again,  5)tV— \  Ans.      fraction  by  2,  it  becomes  -{%• : 

again,  dividing  both  by  5,  we 

obtain  •£,  whose  terms  are  the  lowest  to  which  the  given  fraction 
can  be  reduced. 

Second  Operation.  If  we  divide  both  terms  by   10,   then 

10)i-£— i  Ans.  greatest   common  divisor,  (Art.  170,)  the 

given  fraction  will  be  reduced  to  its  lowts4 
terms  by  a  single  division.     Hence, 

QUEST.— When  the  numerator  is  larger  than  the  denominator,  what  ?  When  smalles, 
what  1  Obs.  What  changes  may  be  made  in  the  terms  of  a  fraction  1  194.  What  la 
meant  by  reduction  of  fractions  1  195.  How  is  a  fraction  reduced  to  its  lowest  terms  1 


112  REDUCTION    OP  [SECT.  VII. 

195»  To  reduce  a  fraction  to  its  lowest  terms. 

Divide  the  numerator  and  denominator  by  any  number  which 
will  divide  them  both  witJwut  a  remainder  ;  and  thus  continue  the 
operation,  till  there  is  no  number  greater  tfian  1  that  will  divide 
them  exactly. 

Or,  divide  b^th  the  i.umerator  and  denominator  by  tlieir  greatest 
common  divisor  ;  the  two  quotients  thence  arising  will  be  the  lowest 
erms  to  which  the  given  fraction  can  be  reduced.  (Art.  170.) 

OBS.  I.  Since  halves  are  larger  than  twentieths,  it  may  be  asked,  how  the 
fraction  £,  can  be  said  to  be  in  lower  terms  than  -£-g-,  It  should  be  observed, 
the  expression  lowest  term,  has  reference  to  the  number  of  parts  into  which  the 
unit  or  thing  is  divided,  and  not  to  the  xalm  or  size  of  the  parts.  Thus,  in  J, 
there  are  fewer  parts  than  in  ^--J-;  in  ^-.  there  are  fewr  parts  than  in  -j^-,  &c. 
Hence,  a  fraction  is  said  to  be  reduced  to  its  l-owest  terms,  when  its  numerator 
and  denominator  are  expressed  in  the  smallest  numbers  possible. 

2.  The  value  of  a  fraction  is  not  altered  by  reducing  it  to  its  lowest  terms  ; 
for,  the  numerator  and  denominator  are  both  divided  by  the  same  number. 

3.  When  the  terms  of  the  fraction  are  small,  the  former  method  will  gen- 
erally be  found  to  be  the  shorter  and  more  convenient  ;  but  when  the  terms 
are  large,  it  is  often  difficult  to  determine  whether  the  fraction  is  in  its  simplest 
form,  without  finding  the  greatest  common  divisor  of  its  terms. 

2.  Reduce  -p-8  to  its  lowest  terms.  Ans.  -£. 

3.  Reduce  A.  11.  Reduce  -?ff. 

4.  Reduce  A-  12.  Reduce  •&*&. 

5.  Reduce  if.  13.  Reduce  ff£. 

6.  Reduce  ff.  14.  Reduce  -fff. 

7.  Reduce  4ff.  15.  Reduce  $&. 

8.  Reduce  ff.  16.  Reduce  if-ff. 

9.  Reduce  -fifr.  17.  Reduce  itf-f. 
10.  Reduce  iVr-  IS.  Reduce  ff  If. 

CASE     II. 

19    Reduce  ^7a  to  a  whole  or  mixed  number. 

Analysis.  —  The  object  in  this  example,  is  to          Operation. 
*h.d  a  whole,  or  mixed  number,  whose  value  is  7)23 

equal   to    the    given   fraction.      Now,   since  7  3f  Ans 


.—Gbs.  What  is  meant  by  the  expression,  lowest  terms  ?  When  is  a  fraction 
said  to  be  reduced  to  its  lowest  terms  ?  Is  the  value  of  a  fraction  altered  by  reducing,  It 
to  its  lowest  terms  ?  Why  not. 


.ARTS.  195-197.  |  FRACTIONS.  113 

sevenths  make  1  whole  one,  23  sevenths  will  make  as  many 
whole  ones  as  7  is  contained  times  in  23.  And  23-r-7  =  3f. 
But  the  value  of  a  fraction  is  the  quotient  of  the  numerator 
divided  by  the  denominator.  (Art.  185.)  Hence, 

196*  To  reduce  an  improper  fraction  to  a  whole,  or  mixed 

number. 

Divide  the  numerator  by  the  denominator,  and  the  quotient  w*H 
be  the  whole,  or  mixed  number  required. 

20.  Reduce  *£•  to  a  whole  or  mixed  number.     Ans.  6*-. 

Reduce  the  following  fractions  to  whole  or  mixed  numbers: 

21.  Reduce  Y-  26.  Reduce  -W- 

22.  Reduce  ^.  27.  Reduce  *£*•. 

23.  Reduce  if.  28.  Reduce 

24.  Reduce  V"-  29.  Reduce 

25.  Reduce  *-$.  30.  Reduce 


CASE    III. 
31.  Reduce-  the  mixed  number  27£  to  an  improper  fraction. 

Operation. 

Analysis.  —  In  1  there  are  5  fifths,  and  in  27  27-£ 

there  are  27  times  as  many.     Now  5X27  =  135,  5 

and  2  fifths  make  137  fifths.     Hence, 


197*  To  reduce  a  mixed  number  to  an  improper  fraction. 

Multiply  the  whole  number  by  the  denominator  of  the  fractiont 
and  to  the  product  add  the  given  numerator.  The  sum  placed  over 
the  given  denominator,  will  form  the  improper  fraction  required. 

OBS.  1.  Any  whole  number  may  be  expressed  in  the  form  of  a  fraction  with- 
out altering  its  value,  by  making  1  the  denominator. 

2.  A  whole  number  may  also  be  reduced  to  a  fraction  of  any  denominator, 
by  multiplying  the  given  number  by  the  proposed  denominates  ;  the  product 
will  be  the  numerator  of  the  fraction  required. 


QUEST.— 196.  How  is  an  improper  fraction  reduced  to  a  whole  or  mixed  number? 
197.  How  reduce  a  mixed  number  to  an  improper  fraction?  Obs.  How  express  a  whole 
number  in  the  form  of  a  fraction  ?  How  reduce  it  to  a  fraction  of  a  given  denominator  1 


114  REDUCTION    OF  [SfiCT.  Vll 


Thus,  25  may  be  expressed  by  *?-,  -HP-,  or  W-,  &c.,  for  2£= 
.^-L^A^  &c>  So  12=^=-^=^=^  for  the  quotient  of 
each  of  these  numerators  divided  by  its  denominator,  is  12. 

32.  Reduce  14£  to  an  impropei  fraction.     Ans.  -^a. 
lleduce  the  following  numbers  to  improper  fractions  : 

33.  Reduce  I7f.  38.  Reduce  856fg-. 

34.  Reduce  25-f.  39.  Reduce  1304£. 

35.  Reduce  48f  40.  Reduce  4725|. 

36.  Reduce  70-^.  41.  Reduce  445  to  tenths. 

37.  Reduce  115-&-.  42.  Reduce  672  to  eighths. 

43.  Reduce  3830  to  one  hundred  and  fifteenths. 

44.  Reduce  5743  to  six  hundred  and  twenty-fifths. 

CASE    IV. 

45.  Reduce  •§  of  •£  to  a  simple  fraction. 

Analysis.  —  f  of  %  is  2  times  as  much  as  1  third  of  -f-.  Now  •§• 
of  -f-  is  g—  3,  or  -fa  ;  for,  multiplying  the  denominator  divides  the 

value  of  the  fraction.  (Art.  188.)     And  2  thirds  is  2  times  2*4,  or 

7x2 

-^-,  which  is  equal  to  if,  or  -&.  (Art.  195.)     The  answer  is  -fo. 

OBS.  This  operation  consists  in  simply  multiplying  the  two  numerators  to- 
gether and  the  two  denominators.  Hence, 

198.  To  reduce  compound  fractions  to  simple  ones. 
Multiply  all  the  numerators  together  for  a  new  numerator,  and 
all  the  denominators  together  for  anew  denominator. 

OBS.  1.  That  a  compound  fraction  may  be  expressed  by  a  simple  one,.  is  evi- 
dent from.  the  fact  that  a  part  of  a  part,  must  be  equal  to  some  part  of  the 
whole. 

2.  The  reason  of  the  rule  may  be  seen  from  the  analysis  of  the  preceding 
example. 

46.  Reduce  -f  of  -f-  of  f  of  -f  to  a  simple  fraction. 

Ans.  T^ph-,  or  ^. 

47.  Reduce  -f  of  i  of  -f^  of  ^  to  a  simple  fraction. 

48.  Reduce  \  of  f  of  f  of  t  of  -fa  to  a  simple  fraction. 

QUEST.—  198.  How  are  compound  fractions  reduced  to  simple  ones? 


ARTS.  198,  199.]  FRACTIONS.  115 

49.  Reduce  f  of  f  of  -fa  of  -^  to  a  simple  fraction. 

50.  Reduce  i  of  f  of  -f  of  f  of  £  to  a  simple  fraction. 

Analysis. — Since  the  product  of  Operation. 

the  numerators  is  to  be  divided  1  #  $  5  &_  6 
by  the  product  of  the  denomina-  jj  $  4  7  a~  &>* 
tors,  we  may  cancel  the  factors  2, 

3,  and  4,  which  are  common  to  both ;  for,  this  is  dividing  the 
terms  of  the  new  fraction  by  the  same  number,  (Art.  148,)  and 
therefore  does  not  alter  its  value.  (Art.  191.)  Multiplying  the 
remaining  factors  together,  we  have  0-\,  which  is  the  answer  re- 
quired. Hence, 

199.  To  reduce  compound  fractions  to  simple  ones  by 
CANCELATION. 

Cancel  all  the  factors  which  arc  common  to  the  numerators  and 
denominators ;  then  multiply  the  rerr^initiy  terms  together  as  be- 
fore. (Art.  198.) 

OBS.  1.  The  reason  of  this  rule  depends  upon  the  fact  that  the  numerator 
and  denominator  of  the  new  fraction  are,  in  effect,  divided  by  the  same  num- 
bers ;  for,  canceling  a  factor  of  a  number  divides  the  number  by  that  factor. 
(Art  148.)  Consequently  the  value  of  the  fraction  is  not  altered.  (Art.  191.) 

2.  This  method  not  only  shortens  the  operation  of  multiplying,  but  at  the 
same  time  reduces  the  answer  to  its  lowest  terms.  A  little  practice  will  give 
the  student  great  facility  in  its  application. 

51.  Reduce  f  of  -£-f  of  f  to  a  simple  fraction. 

Operation. 

3  First  we  cancel  the  3  and  8  in  the 

$       1$    f  $__3     .  numerator,  then  the  24  in  the  denomina- 

0       $£       7~~7  tor,  which  is  equal  to  :he  factors  3  into  8. 

Finally,  we  cancel  the  5  in  the  denomina- 
tor and  the  factor  5  in  the  numerator  15,  placing  the  other  factor 
3  above.  We  have  3  left  in  the  numerator,  and  7  in  the  denom 
inator.  Ans.  f. 

52.  Reduce  -f  of  -f  of  -f  of  if  to  a  simple  fraction. 

53.  Reduce  |  of  ^  of  |  of  -fa  of  f  to  a  simple  fraction. 

QUEST.— 199.  How  by  cancelation  ?    How  does  it  appear  that  this  method  will  gtaj  th« 
true  answer  ?     Obs.  What  advantages  does  this  method  possess  1 
T.H. 


116  REDUCTION    OF  fbfiCT.  VIL 

54.  Roduce  •§•  of  f  of  -f  of  -f-  of  -ft-  to  a  simple  fr  fiction. 

55.  Reduce  f  of  -fc  of  -f  of  •§•  £  to  a  simple  fraction. 

56.  Reduce  •£  of  ^-f  of  f  of  3*5-  to  a  simple  fraction. 

57.  Reduce  f  of  -tf  of  -ff  of  -J-g-  to  a  simple  fraction. 

58.  Reduce  -f-  of  -ft-  of  -ft-  of  f  of  -f  to  a  simple  fraction. 
59    Reduce  i  of  f  of  4f  of  f  of  -f-  to  a  simple  fraction. 
60.  Reduce  f  of  3^  of  f  of  i^  of  -£  to  a  simple  fraction. 
Note. — For  reduction  of  complex  fractions  to  simple  ones,  see  Art.  239 

CASE    V. 

Ex.  61.  Reduce  ^  ana  -J-  to  a  common  denominator. 

Aofe. — Two  or  more  fractions  are  said  to  have  a  common  denominator,  when 
they  have  the  same  denominator. 

Solution. — If  both  terms  of  the  first  fraction  -fc,  are  multi- 
plied by  the  denominator  of  the  second,  it  becomes  -fa ;  and  if 
both  terms  of  the  second  fraction  -J-,  are  multiplied  by  the  de- 
nominator of  the  first,  it  becomes  -ft-.  Thus  the  fractions  -fa  and 
ft-  have  a  common  denominator,  and  are  respectively  equal  to  the 
given  fractions,  viz :  -fa=i,  and  T^=-}-.  (Art.  191.)  Hence, 

2OO.  To  reduce  fractions  to  a  common  denominator. 

Multiply  each  numerator  into  all  the  denominators  except  its 
own  for  a  new  numerator,  and  all  tJie  denominators  together  for  a 
common  denominator. 

62.  Reduce  \,  f ,  and  f  to  a  common  denominator. 

Operation. 
1X4X6=24  \ 

3  X  3  X  6=54  >  the  tliree  numerators. 
5X3X4=60  ) 
3X4X6=72  the  common  denominator. 

Ans.  ff,  ft,  and  f £. 

OBS.  The  reason  that  the  process  of  reducing  fractions  to  a  common  denom 
inator  does  not  alter  their  value,  is  because  the  numerator  and  denominator  ol 
each  of  the  given  fractions,  are  multiplied  by  the  same  numbers ;  and  multiplying 

QUEST.— Note.  What  is  meant  by  a  common  denominator  7  200.  How  are  fractions  re- 
dnced  to  a  common  denominator  ?  Obs.  Does  the  [focess  of  reducing  fractions  to  a  com 
moa  denominator  alter  their  value  1  Why  not  ? 


ARTS.  200,  201.]  FRACTIONS.  117 

both  tho  numerator  and  denominator  of  a  fraction  by  the  same  number,  doe» 
not  alter  its  value.  (Art.  191.) 

63.  Reduce  f,  f,  -J-,  and  -f-  to  a  common  denominator. 

64.  Reduce  f ,  i,  f,  and  f  to  a  common  denominator. 
Reduce  the  following  fractions  to  a  common  denominator: 

65.  Reduce  f,  i,  £,  and  -f.  69.  Reduce  if,  ft,  and  ffr. 

66.  Reduce  f ,  -f,  -f,  and  f .  70.  Reduce  if,  -j2^,  and  ff. 

67.  Reduce  f,  -f,  -ft,  and  T^/      71.  Reduce  ff,  •§-*,  and  i£. 

68.  Reduce  ft,  f,  if,  and  f.       72.  Reduce  ££,  f£,  and  iff- 

CASE    VI. 
73.  Reduce  •£,  -f-,  and  -|  to  the  least  common  denominator. 

Analysis. — We  first  find  the  least          Operation. 
common  multiple  of  all  the  given  de-      2)3  "  4  "  8 
nominators,  which  is  24.  (Art.  176.)       2)3  "  2  "  4 
The  next  step  is  to  reduce  the  given          3  "  1  "  2 
fractions   to   twenty-fourths  without     Now  2X2X3X2  =  24,  the 
altering  their  value.     This  may  evi-     least  common  denominator, 
dently  be  done  by  multiplying  both 

terms  of  each  fraction  by  such  a  number  as  will  make  its  denom- 
inator 24.  (Art.  191.)  Thus  3,  the  denominator  of  the  first  frac- 
tion, is  contained  in  24,  8  times ;  now,  multiplying  both  terms  of 
the  fraction  i  by  8,  it  becomes  /f.  The  denominator  4,  is  con- 
tained in  24,  6  times  ;  hence,  multiplying  the  second  fraction  f  by 
6,  it  becomes  if .  The  denominator  8,  is  contained  in  24,  3  times  : 
and  multiplying  the  third  fraction  f  by  3,  it  becomes  if.  Ther«- 
fore  2\,  if,  and  if  are  the  fractions  required.  Hence, 

20 1.  To  reduce  fractions  to  their  least  common  denominator. 

I.  Find  the  least  common  multiple  of  all  the  denominators  of 
the  given  fractions,  and  it  will  be  the  least  common  denominator, 
(Art.  176.) 

II.  Divide  the  least  common  denominator  by  the  denominate 
cf  each  given  fra.ctlm,  and  multiply  the  quotient  by  the  numerator  * 
the  products  will  be  tlie  numerators  of  the  fractions  required. 

GniflT  — 201.  How  are  fractions  reduced  to  the  least  common  denominator  I 


118  REDUCTION    OF  [SECT.   VII, 

OBS.  1.  This  process,  in  effect,  multiplies  both  the  numerator  f.nd  denomina- 
tor of  the  given  fractions  by  the  same  number,  and  consequently  does  not  alte 
their  value  (Art.  191.) 

2.  The  rule  supposes  each  of  the  given  fractions  to  be  reduced  to  its  lowest 
terms ;  otherwise,  the  least  common  multiple  of  their  denominators  may  not  be 
the  least  common  denominator  to  which  the  given  fractions  are  capable  of  being 
reduced.     Thus,  the  fractions  ^,  f-,  and  -^,  when  reduced  to  the  least  com- 
mon denominator  as  they  stand,  become  -fa,  -j6^,  and  -j9^.     But  it  is  obvious 
that  these  fractions  are  not  reduced  to* their  least,  common  denominator;  for, 
fchey  can  be  reduced  to  1,  |-,  and  |k     Now,  if  the  given  fractions  are  reduced 
to  the  lowest  term*,  they  become  $,  £,  and  |,  and  the  leasl  wmmon  multiple  of 
their  denominators,  is  also  4.  (Art.  176.) 

3.  By  a  moment's  reflection  the  student  will  often  discover  the  least  common 
den  -aiinator  of  the  given  fractions,  without  going  through  the  ordinary  pro- 
ce»  of   finding  the  least  common  multiple  of  their  denominators.     Take  the 
fra  Hons  i,  f ,  and  -fa  ;    the  least  common  denominator,  it  will  be  seen  at  a 
gl?  >ce,  is  4.     Now  if  we  multiply  both  terms  of  i  by  2,  it  becomes  -| ;  and  if 
we   livide  both  terms  of  -fa  by  3,  or  reduce  it  to  its  lowest  terms,  it  becomes  {. 
TJ   is  the  given  fractions  are  equal  to  |,  |,  and  i,  and  are  reduced  to  the  least 
co    man  deiwmiuator. 

74.  Reduce  -f,  f-,  and  •£  to  the  least  common  denominator. 

Operation.  Now  2X2X3X2  =  24,  the  least  com.  denom. 
2)4  "  6  "  8  Then  24-4  =  6,  and  6X3  =  18,  the  1st  num. 
2)2  "  3  "  4  24  — 6  =  4,  and  4X5  =  20,  the  2d  " 

1  "  3~/r2  24  — 8  =  3,  and  3X7  =  21,  the  3d     " 

Ans.  if,  fi,  and  ift. 

T5.  Reduce  -f-  and  -fa  to  the  least  common  denominator. 
Reduce  the  following  fractions  to  the  least  common  denominator: 

76.  A,  i,  i,  and  -fa,  84.  if,  H,  if,  and  ffj. 

77.  |,  f,  and  i.  85.  ^  H>  it,  and  /2t 

78.  i,  |,  1,  and  H.  86.  A,  ii,  if  and  -ff. 

79.  f,  f,  f,  and  -fV-  87.  if,  ffr,  ^  and  f£. 

80.  f ,  i,  f,  and  if.  88.  ff ,  if,  ff,  and  if. 

81.  i,  1*0,  iHK  and  ffr.  89.  fi,  i£,  if,  and  ££. 

82.  -&,  -H-,  T^,  and  ff.  90.  fft-,  if,  fi  and  -ffi-. 

83.  ^,  I,  if,  and  ii.  91.  fi,  -W,  ti,  and  ^. 


*iuEST.— 9As.  Does  this  process  alter  the  value  of  the  Riven  fractions?    Why  ootf 
rhat  does  this  rule  suppose  respecting  the  given  fractions  7 


ARTS.  201, 202.  J  FRACTIONS.  119 

ADDITION   OF  FRACTIONS. 

Ex.  1.  A  beggar  meeting  four  persons,  obtained  f  of  a  dollar 
from  the  first,  •£•  from  the  second,  |-  from  the  third,  and  £  from 
the  fourth :  how  much  did  he  receive  from  all  ? 

Solution. — Since  the  several  donations  are  all  in  the  same  pans 
of  a  dollar,  viz :  sixths,  it  is  plain  they  may  be  added  together  in 
the  same  manner  as  whole  dollars,  whole  yards,  &c.  Thus,  1 
sixth  and  3  sixths  are  4  sixths,  and  4  are  8  sixths,  and  5  are  13 
sixths.  Ans.  Jga,  or  2£  dollars. 

Ex.  2.  What  is  the  sum  of  -f  and  f-  ? 

OBS.  A  difficulty  here  presents  itself  to  the  learner ;  for,  it  is  evident,  that 
2  thirds  and  3  fourths  neither  make  5  thirds,  nor  5  fourths.  (Art.  51.)  This 
difficulty  may  be  removed  by  reducing  the  given  fractions  to  a  common  de- 
nominator. (Art.  200.)  Thus, 

Operation. 

2X4  =  8 


_„  t  the  new  numerators. 
—  y 

3X4=12,  the  common  denominator. 

The  fractions,  when  reduced,  are  i\  and  T\;  now  8  twelfths  X 
9  twelfths=l7  twelfths.  Ans.  -j-f,  or  IT\. 

2 O2.  From  these  illustrations  we  deduce  the  following  general 

RULE  FOR  ADDITION  OF   FRACTIONS. 

Reduce  the  fractions  to  a  common  denominator ;  add  their  nu- 
merators, and  place  t/ie  sum  over  the  common  denominator. 

OBS.  1.  Cmnpmind  fractions  must,  of  course,  be  reduced  to  simple  ones,  be- 
fore attempting  to  reduce  them  to  a  common  denominator.  (Art.  198.) 

2.  Mixed  numbers  may  be  reduced  to  improper  fractions,  and  then  be  added 
according  to  the  rule;  or,  we  may  add  the  whole  numbers  and  fractional  parte 
separately,  and  then  unite  their  sums. 

3.  In  many  instances  the  operation  may  be  shortened  by  reducing  the  give» 
fractions  to  the  least  common  denominator.  (Art.  120  ) 

QUEST. — 202.  How  are  fractions  added?  Obs.  What  must  be  done  with  ccmpouni 
fractions  ?  How  are  mixed  n  'inbers  added  ?  How  may  the  operation  frequently  be  short- 
ened i 


120  ADDITION    OF  [SECT.  VII 

EXAMPLES. 

3.  What  is  the  sum  of  i,  £,  and  f-  ?     Ans.  V=2. 

4.  What  is  the  sum  of  i,  -f ,  -f ,  and  f  ? 

5.  What  is  the  sum  of  f ,  f ,  i,  and  f  ? 

6.  What  is  the  sum  of  f,  f ,  If,  and  £  ? 

7.  What  is  the  sum  of  -f,  i%,  f ,  and  -&  ? 

8.  What  is  the  sum  of  f,  f ,  ft,  and  A  ? 

9.  What  is  the  sum  of  f,  -fV,  f,  and  f  ? 

10.  What  is  the  sum  of  f,  -fo  f,  and  V? 

11.  What  is  the  sum  of  i,  -J-,  i,  -f,  and  f  ? 

12.  What  is  the  sum  of  f  of  i,  f  of  f ,  and  f  ? 

13.  What  is  the  sum  of  i  of  f,  -f  of  £,  and  -ft  ? 

14.  What  is  the  sum  of  f-  of  |  of  $  of  i,  and  |  ? 

15.  What  is  the  sum  of  f ,  -f  of  3,  f  of  i,  and  i  ? 

16.  What  is  the  sum  of  4|,  8i,  2£,  6f,  and  |? 

17.  What  is  the  sum  of  i  of  6,  f  of  2,  3£,  and  5?  ? 

18.  What  is  the  sum  of  f,  if,  ffr,  ii,  and  -j-$  ? 

19.  What  is  the  sum  of  2l£,  35i,  -fi,  and  f  of  1? 

20.  What  is  the  sum  of  i  of  f ,  ^,  6|,  If,  and  -|  ? 

21.  What  is  the  sum  of  i  and  1*4  ? 

Afofe. — It  is  obvious,  if  two  fractions,  each  of  whose  numerators  is  1,  are  re* 
duced  to  a  common  denominator,  the  new  numerators  will  be  the  same  as  the 
given  denominators.  (Art.  200.)  Thus,  if  -|  and  -fa  are  reduced  to  a  common 
denominator,  the  new  numerators  will  be  1 2  and  8,  the  same  as  the  given  de- 
nominators. Now.  the  sum  of  the  new  numerators,  placed  over  the  product 

1*2— UH     20 
of  the  denominators,  will  be  the  answer;  (Art.  202;)  that  is  TS-T^=QT:»  or 

l^Xo    •'o 
2?- ,  the  answer  required.     Hence, 

2O3«  To  find  the  sum  of  any  two  fractions  whose  numerators 
are  one. 

Add  tJie  denominators  together,  place  this  sum  over  their  prod- 
uct, and  tlue  result  will  be  the  answer  required. 

OBS.  1.  The  reason  of  this  rule  may  be  seen  from  the  fact  that  the  opera 
lion  is  the  same  as  reducing  the  given  fractions  to  a  common  denominator 
then  adding  their  numerators. 

2.  When  the  numerators  of  two  factors  are  the  same,  their  sum  may  be  round 


QUEST.— 303.  How  is  the  sum  of  any  two  fractions  found  whose  numerators  are  1 1 
Obs.  How  find  the  sum  of  two  fractions  whose  numerators  are  the  same  1 


ARTS.  203,  204.]  FRACTIONS.  121 

by  multiplying  the  sum  of  the  two  denominators  hy  the  common  numerator, 
and  placing  the  result  over  the  product  of  the  given  denominators      Thus,  the 

,     (4-J-5)  X3    9X3    27       ,  - 
sum  of  -3-  and  f  is  equal  to       ^^—-^———^  or  1-jfr. 


22.  What  is  the  sum  of  -2-V  and  -gV?  Of  jV  and  6-*§  ? 

23.  What  is  the  sum  of  -5-V  and  ^  ?  Of  •&  and  -§V  ? 

24.  What  is  the  sum  of  -£g  and  gV  ?  Of  -J^T  and  -g-ir  ? 

25.  What  is  the  sum  of  -|  and  -fa  ?  Of  -&  and  -&  ? 
20    What  is  the  sum  of  ^  and  -fl  ?  Of  £*  and  if  ? 

27.  What  is  the  sum  of  if  and  ffr  ?  Of  ffr  and  -^ft-  ? 

28.  What  is  the  sum  of  5  and  -f  ? 

Note.  —  The  design  in  tliis  and  the  following  examples,  is  to  incorporate  the 
integers  with  the  fractions,  and  express  the  answer  fractionally. 

Solution.  —  5=-^-.    (Art.  197.  Obs.  2.)     Now  -^-+f  =Y  Ans. 
2O  4.  Hence,  to  add  a  whole  number  and  a  fraction  together. 

O 

Reduce  the  whole  member  to  a  fraction  of  the  same  denominator 
as  that  of  the  given  fraction  ;  then  add  their  numerators  together. 
(Arts.  202,  197.  Obs.'  1,  2.) 

Nvtc.  —  The  process  of  incorporating  a  whole  number  with  a  fraction,  is  the 
same  as  that  of  reducing  a  mixed  number  to  an  improper  fraction.  (Art.  197.^ 

29.  What  is  the  sum  of  45  and  f  ? 

30.  What  is  the  sum  of  320  and  -fc  ? 

31.  What  is  the  sum  of  452  and  VW? 

32.  What  is  the  sum  of  6  3  5H  +4  2  7^+16251? 

33.  What  is  the  sum  of  195H+600lHH-5630!-H-160£? 

34.  What  is  the  sum  of  67ltl+483it+8421^-f  4325i? 

35.  What  is  the  sum  of  590ii+100fi+  4005|f  +3020^? 

36.  What  is  the  sum  of  239|f  +644|i+1650|f  +4500tV? 
37    What  is  the  sum  of  6563i+1000i+1830f  +83001? 

38.  What  is  the  sum  of  356H+46f+105i+000|+321-|  ? 

39.  What  is  the  sum  of  41^-f  105|-h300f-f  241f  +472-J-? 

40.  What  is  tin  sum  of  8G72!-f-163G45i-f-lSOOi-}-GG251fV? 

41.  What  is  the  sum  of  2G003|+19352-f  +  92S31-f-68G93f  ? 

42.  What  is  the  sum  of  1  9256-^+456001+1  of  f  of  f  ? 

43.  What  is  the  sum  of  f  of  28  +  6-^+45^  +£  of  300  ? 

QUEST.  —  204.  How  add  a  whole  number  ami  a  fraction  ? 


122  SUBTRACTION    OF  [bECT. 

SUBTRACTION   OF   FRACTIONS. 

2O  5  •  Ex.  1.  A  man  bought  -^  of  an  acre  of  land,  and  after- 
"wards  sold  -^  of  it  :  bow  much  land  had  he  left  ? 

Solution.  —  7  tenths  from  9  tenths  leave  2  tenths. 

Am.  -fs  of  an  acre. 

2    A  laborer  having  received  -£  of  a  dollar  for  a  day's  work., 
pent  £  of  a  dollar  for  liquor  :  how  much  money  had  he  left  ? 

Note.  —  The  learner  meets  with  tHfe  same  difficulty  here  as  in  the  second  ei' 
ample  of  adding  fractions;  that  is,  he  can  no  more  subtract  fifflis  from  eighths^ 
than  he  can  add  fifths  to  eighths;  for,  •§•  of  a  dollar  taken  from  -J  of  a  dollar  will 
neither  leave  4  fifths,  nor  4  eighths.  The  fractions  must  therefore  be  reduced 
to  a  common  denominator  before  the  subtraction  can  be  performed. 

Operation, 

3X8  =  24  (  ^e  numerators-  (Art-  200.) 
8X5  =  40,  the  common  denominator. 
The  fractions  become  -f-ft  and  -££.     Now  £{j  —  H=ii 


2OG»  From  these  illustrations  we  deduce  the  following  general 

RULE  FOR  SUBTRACTION  OF  FRACTIONS. 

Reduce  the  given  fractions  to  a  common  denominator  ;  subtract 
the  less  numerator  from  the  greater,  and  place  the  remainder  over 
the  common  denominator. 

OBS.  Compound  fractions  must  be  reduced  to  simple  ones,  as  in  addition  of 
fractions.  (Art.  198.) 

EXAMPLES. 

3.  From  -f  take  ^.  Ans.  •/•$. 

4.  From  -ff  take  A-  9-  'From  ff  take  ff  . 

5.  From  -ff  ^ke  if.  10.  From  -f  of  f  take  |  of  f. 

6.  From  if  take  f.  11.  From.f  of  i  take  i  of  ^. 

7.  From  H  take  |f.  12.  From  |  of  40  take  f  of  20. 

8.  From  ff  take  if.  13.  From  f  of  f  of  i  take  f  of  £ 

QUEST.—  206.  How  is  one  fraction  subtracted  from  another  7  Ofs.  Wtat  is  to  be  orone 
with  compound  fraction*  ? 


ARTS.  205-208.]  FRACTIONS.  123 

2O7.  Mixed  numbers  may  be  reduced  to  improper  fractions, 
then  to  a  common  denominator,  and  be  subtracted  ;  or,  the  frac- 
tional part  of  the  less  number  may  be  taken  from  the  fractional 
part  of  the  greater,  and  the  less  whole  number  from  the  greater. 

14.  From  9-J-  take  7^. 

First  Operation.  Second  Operation. 


Ans.  i=li,  or  l£.  Ans.  If,  or  1£. 

Note.  —  Since  we  cannot  take  3  fourths  from  1  fourth,  we  borrow  a  unit  jn 
the  second  operation  and  reduce  it  to  fourths,  which  added  to  the  1  fourth, 
make  5  fourths.  Now  3  fourths  from  5  fourths  leave  2  fourths  :  1  to  carry  to 
7  makes  8,  and  8  from  9  leaves  1. 

15.  From  25f  take  13-f.  17.  From  178^  take  56-f. 

16.  From  230-22S-  take  160-fV.       18.  From  761fi  take  482-^ 

19.  From  5  take  f. 

Suggestion.  —  Since  3  thirds  make  a  whole  one,  in  5  whole  ones 
there  are  15  thirds;  now  2  thirds  from  15  thirds  leave  13  thirds. 
Ans.  -f,  or  4^.  Hence, 

2O  8.  To  subtract  a  fraction  from  a  whole  number. 

Change  the  whole  number  to  a  fraction  having  the  same  denom- 
inator as  the  fraction  to  be  subtracted,  and  proceed  as  before. 
(Art.  197.  Obs.  2.) 

OBS.  If  the  fraction  to  be  subtracted  is  a  proper  fraction,  we  may  simply 
borrow  a  unit  and  take  the  fraction  from  this,  remembering  to  diminish  the 
whole  number  by  1.  (Art.  G9.  Obs.  1.) 

20.  From  20  take  f.  ^^ns.  19f. 

21.  From  135  take  9|.  26.'  From  720  take  125^. 

22.  From  263  take  24+£-.  27.  From  1000  take  25  ft. 

23.  From  168  take  30-|.  28.  From  563  take  562f|. 

24.  From  567  take  lOOffr.  29.  From  9263  take  999-J-. 

25.  From  634  take  342f.  30.  From  857  take  785  ii. 


Q.UKST. — 207.  How  are  mixed  numbers  subtracted?    208.  How  is  a  fraction  subtracter 
from  a  whole  number  ? 
6* 


124  MULTIPLICATION    OP  [Sp.CT.  VII. 

MULTIPLICATION  OF   FRACTIONS. 

2O 9.  We  have  seen  that  multiplying  by  a  whole  number,  is 
taking  the  multiplicand  as  many  times  as  there  are  units  in  the 
multiplier.  (Art.  82.)  On  the  other  hand, 

If  the  multiplier  is  only  a  part  of  a  unit,  it  is  plain  we  must 
take  only  a  part  of  the  multiplicand.  That  is, 

Multiplying  by  £,  is  taking  1  half  of  the  multiplicand  once. 
Thus,  ]2Xi=6. 

Multiplying  by  •£,  is  taking  1  third  of  the  multiplicand  once. 
Thus,  12xi=4. 

Multiplying  by  f,  is  taking  1  third  of  the  multiplicand  twice. 
Thus,  12  X -3=8.  Hence, 

2 1  O.  Multiplying  by  a  fraction  is  taking  a  certain  PORTION 
of  the.  multiplicand  as  many  times,  as  there  are  like  portions  of  a 
unit  in  the  multiplier. 

OBS.  If  the  multiplier  is  a  unit  or  1,  the  product  is  equal  to  the  multiplicand  ; 
if  the  multiplier  is  greater  than  a  unit,  the  product  is  greater  than  the  multi- 
plicand ;  (Art.  82 ;)  and  if  the  multiplier  is  less  than  a  unit,  the  product  is 
less  than  the  multiplicand. 

CASE    I. 

211.   To  multiply  a  fraction  and  a  whole  number  together. 
Ex.  1 .  If  1  man  drinks  f  of  a  barrel  of  cider  in  a  month,  how 
much  will  5  men  drink  in  the  same  time  ? 

Analysis. — Since  1  man  drinks  ^  of  a  barrel,  5  men  will  drink 
6  times  as  much;  and  5  times  2  thirds  are  10  thirds;  that  is, 
fX5=Y>  or  3|.  (Art.  196.)  Ans.  3£  barrels. 

Ex.  2.  If  a  pound  of  tea  costs  |  of  a  dollar,  how  much  will 
i  pounds  cost  ? 

Solution. — 1x4=-^-;  Imd  ^^2-f,  or  2i  dons.  Ans. 
Or,  since  dividing  the  denominator  of  a  fraction  by  any  num 
bei  multiplies  the  value  of  the  fraction  by  that  number,  (Art.  189,) 

QUEST.— 209.  What  is  meant  by  multiplying  by  a  whole  number  ?  210.  What  is  meant 
•y  multiplying  by  a  fraction  ?  Obs.  If  the  multiplier  is  a  unit  or  1,  what  is  the  product 
equal  to  ?  When  the  multiplier  is  greater  than  1,  how  is  the  product,  compared  with  th« 
multiplicand  ?  When  less,  how  ? 


ARTS.  209-21  3.  J  FRACTIONS.  125 

if  we  divide  the  denominator  8  by  4,  the  fraction  will  become  f  , 
which  is  equal  to  2-£,  the  same  as  before.     Hence, 


To  multiply  a  fraction  by  a  whole  number. 
Multiply  the  numerator  of  the  fraction  by  tlie  whole  number, 
and  write  the  product  over  the  denominator. 

Or,  divide  the  denominator  by  the  whole  number,  when  this  can 
be  done  without  a  remainder.  (Art.  180.) 

OBS.  1.  A  fraction  is  multiplied  into  a  number  equal  to  its  denominator  by 
canceling  the  denominator.  (Ax.  9.)  Thus  ^X"7=4. 

2.  On  the  same  principle,  a  fraction  is  multiplied  into  any  factor  m  its  de- 
<n0miualf>r,  by  canceling  that  factor.  (Art.  189.)     Thus,  -j^xS—  f  . 

3.  Since  multiplication  is  the  repeated  addition  of  a  number  or  quantity  to 
ilsc/f,  (Art.  HO,)  the  student  sometimes  finds  it  difficult'to  account  for  the  fact 
that  the  product  of  a  number  or  quantity  by-  a  proper  fraction,  is  always  kss 
than  the  number  multiplied.     This  difficulty  will  at  once  be  removed  by  re- 
flecting that  mv.lt  if  tlyin^  by  a  fraction  is  taJring  or  repeating  a  certain  portion 
of  the  multiplicand  as  many  times,  as  there  are  like  portions  of  a  unit  in  the 
multiplier.  (Art.  '210.) 

EXAMPLES. 

3.  Multiply  -f-Jr  by  15.  Ans.  **&*-,  or  10£. 

4.  Multiply  i£  by  8.  9.  Multiply  fj-  by  165. 

5.  Multiply  ££  by  30.  10.  Multiply  fff  by  100. 

6.  Multiply  |f  by  27.  11.  Multiply  /3\  by  530. 

7.  Multiply  -HHr  by  45.-  12.  Multiply  If  by  1000. 

8.  Multiply  fi  by  100.  13.  Multiply  Hi  by  831. 
14.  Multiply  12|  by  8. 

Operation. 

12  f  8  times  -f  are  *£-,  which  are  equal  to  5  and  -J-. 

8  Set  down  the  i.    8  times  12  are  96,  and  5  (which 

Ans.  101^.  arose  from  the  frac^Hf  make  101.     Hence, 

213.  To  multiply  a  mixed  number  by  a  whole  one. 
Multiply  the  fractional  part  and  the  whole  number  separately 
and  unite  the  products. 

QUEST.—  212.  How  multiply  a  fraction  by  a  whole  number  ?  Ols  How  is  a  fraction 
multiplied  by  a  number  equal  to  its  denominator  ?  How  by  any  factol  la  its  denominator  * 
213.  How  is  a  mixed  number  multiplied  by  a  whole  one  ? 


126  MULTIPLICATION    OF  [SECT.   VII 

15.  Multiply  45i  by  10.  Ans.  4511. 

16.  Multiply  811  by  9.  19.  Multiply  127|  by  35. 

17.  Multiply  31  f£  by  20.  20.  Multiply  48^  by  47. 

18.  Multiply  148-J1  by  25  21.  Multiply  250^  by  50. 

2 1  4.  Multiplying  by  a  fraction,  we  have  seen,  is  taking  a 
certain  portion  of  the  multiplicand  as  many  times,  as  there  are 
like  portions  of  a  unit  in  the  multiplier.  Hence, 

To  multiply  by  •£ :  Divide  the  multiplicand  by  2. 

To  multiply  by  -^ :  Divide  the  multiplicand  by  3. 

To  multiply  by  •£- :  Divide  the  multiplicand  by  4,  &c. 

To  multiply  by  -f :  Divide  by  3,  and  multiply  the  quotient  by  2. 

To  multiply  by  1 :  Divide  by  4,  and  multiply  the  quotient  by  3. 

215.  Hence,  to  multiply  a  whole  number  by  a  fraction. 

Divide  the  multiplicand  by  the  denominator,  and  multif>ly  the 
quotient  by  the  numerator. 

Or,  multiply  the  given  number  by  the  numerator,  and  divide  the. 
product  by  the  denominator. 

Obs.  1.  When  the  given  number  cannot  be  divided  by  the  denominator 
without  a  remainder,  the  latter  method  is  generally  preferred. 

2.  Since  the  product  of  any  two  numbers  is  the  same,  whichever  is  taken 
for  the  multiplier,  (Art  83,)  the  fraction  may  be  taken  for  the  multiplicand, 
and  the  whole  number  for  the  multiplier,  when  it  is  more  convenient. 

22.  If  1  ton  of  hay  costs  21  dollars,  how  much  -will  f  of  a  toE 
cost? 


Analysis.  —  Since  1  ton  costs  21  dollars,  ^  of  4)21 

a  ton  will  cost  -f-  as  much.     Now,  1  fourth  of  21  5 

is  -Vs  and  i  of  21  is  Senes  as  much;  but  3 


s  Scen 
f  cHlar 


—  X3=—  j—  =-^-,  or  15f  clars.  Ans.  15f  dolK 

23.  Multiply  136  by  i.  Ans.  45  •£. 

24.  Multiply  432  by  \.  26.  Multiply  360  by 

25.  Multiply  635  by  £.  •  27.  Multiply  580  by 


UDEST. — 215.  How  is  a  whole  number  multiplied  ly  a  fraction  I    216.  How  find  a  ftae 
tional  part  of  a  number  1 


ARTS.  21J-217.J  FRACTIONS.  127 

28.  Multiply  672  by  f.  31.  Multiply  660  by  ^-. 

29.  Multiply  710  by  f.  32.  Multiply  840  by  $-$. 

30.  Multiply  765  by  H-  33.  Multiply  975  by  iff. 

216.  Since  multiplying  by  a  fraction  is  taking  a  certain  por- 
tion of  the  multiplicand  as  many  times,  as  there  are  like  portions 
of  a  unit  in  the  multiplier,  it  is  plain,  that  the  process  of  finding 
a  fractional  part  of  a  number,  is  simply  multiplying  the  number 
by  the  given  fraction,  and  is  therefore  performed  by  the  same  rule. 

Thus,  -f  of  12  dollars  is  the  same  as  the  product  of  12  dollars, 
multiplied  by  | ;  and  12Xi=8  dollars. 

OBS.  The  process  of  finding  a  fractional  part  of  a  number,  is  often  a  source 
of  confusion  and  perplexity  to  the  learner.  The  difficulty  arises  from  the 
erroneous  impression  that  finding  a  fractional  part,  implies  that  the  given  num- 
ber must  be  divided  by  the  fraction,  instead  of  being  multiplied  by  it. 

34.  What  is  -&  of  457  ?  Ans.  26G-&. 

35.  What  is  if  of  16245  ?  38.  What  is  iff  of  5268  ? 

36.  What  is  £ f-  of  25000  ?  39.   What  is  -fff-  of  45260  ? 

37.  What  is  •£&  of  4261  ?  40.  What  is  -jVoV  of  452120  * 

41.  Multiply  64  by  5£. 

Operation. 

2)64  We  first  multiply  64  by  5,  then  by  £,  and  the 

5£  sum  of  the  products  is  352.     But  multiplying  by 

320  •£  is    taking 'owe   half  of  the   multiplicand  once. 

32  (Arts.  82,  214.)     Hence, 
Ans.  3~52? 

217*  To  multiply  a  whole  by  a  mixed  number. 
Multiply  first  by  the  integer,  then  by  the  fraction,  and  add  the 
products  together.  (Art.  214.) 

42.  Multiply  83  by  7£.  Ans.  597f. 

43.  Multiply  45  by  8£. .  47.  Multiply  225  by  30$. 

44.  M  iltiply  72  by  10^.  48.  Multiply  342  by  20f. 

45.  Multiply  93  by  12f.  49.  Multiply  432  by  35$. 

46.  Multiply  184  by  18|.  50.  Multiply  685  by  42*. 

QIT*ST.— 217.  How  is  a  whole  number  multiplied  by  a  mixed  number  7 


128  MULTIPLICATION    OF  [SECT.  VII 

51    Multiply  125  by  10-2\.  56.  Multiply  457  by  12f£-. 

52.  Multiply  26  by  10££  57.  Multiply  107  by  47|i. 

53.  Multiply  256  by  17-ft-.  58.  Multiply  510  by  85fjf-. 

54.  Multiply  196  by  41£i.  59.  Multiply  834  by  89-fV 
65.  Multiply  341  by  30^.  60.  Multiply  963  by  95fi. 

CASE    II. 

218.  To  multiply  a  fraction  ly  a  fraction. 

Ex  1.  A  man  bought  f  of  a  bushel -of  wheat,  at  1  of  a  dolluf 
per  bushel  •  how  much  did  he  pay  for  it  ? 

Analysis. — Since  1  bushel  costs  £  of  a  dollar,  i  of  a  bushel 
must  cost  i  of  -£,  which  is  4^  of  a  dollar;  for,  multiplying  the 
denominator,  divides  the  value  of  the  fraction.  (Art.  188.)  Now, 
if  -£-  of  a  bushel  costs  -fa  of  a  dollar,  f  of  a  bushel  will  cost  4  times 
as  much ;  and  4  times  -fa  are  f-g-,  or  -ft  dolls.  (Art.  195.) 

Ans.  y7^  of  a  dollar. 

Or,  we  may  reason  thus :  since  1  bushel  costs  •£  of  a  dollar, 
f  of  a  bushel  must  cost  £  of  $  of  a  dollar.  Now  -§-  of  i  is  a  com- 
pound fraction,  whose  value  is  found  by  multiplying  the  numera- 
tors together  for  a  new  numerator,  and  the  denominators  for  a 
new  denominator.  (Art.  198.) 

Solution. — f-Xf =-H,  or  -fo  dollars,  Ans.     Hence, 

219.  To  multiply  a  fraction  by  a  fraction. 

Multiply  the  numerators  together  for  a  new  numerator,  and  the 
denominators  together  for  a  new  denominator. 

OBS.  1.  It  will  be  seen  that  the  process  of  multiplying  one  fraction  by  an- 
other, is  precisely  the  same  as  that  of  reducing  compound  fractions  to  simple 
ones.  (Art.  198.) 

2.  The  reason  of  this  rule  may  be  thus  explained.  Multiplying  by  a  fractica 
is  taking  a  certain  part  of  the  multiplicand  as  many  times,  as  there  are  like 
p(*,r(s  of  a  unit  in  the  multiplier.  (Art.  210.)  Now  multiplying  the  denomina- 
tor of  the  multiplicand  by  the  denominator  of  the  multiplier,  gives  the  vjJue 
if  only  one  of  the  parts  denoted  by  the  given  multiplier;  (Art.  188;)  we  there- 
fore multiply  this  new  product  by  the  numerator  of  the  multiplier,  to  find  the 
number  of  parts  denoiel  by  the  given  multiplier.  (Art.  18G.) 

QUEST.— 219.  How  is  a  fraction  multiplied  by  a  fraction  1  Obs.  To  what  is  the  proceM 
of  multiplying  one  fraction  b>  another  similar? 


ARTS.  218-220.]  FRACTIONS.  129 


2.  Multiply  i  by  f  .  Am.  -fa=-\. 

3.  Multinjy  -f  by  -fr.  6.  Multiply  -f*  by 

4.  Mul^Pii  by  if.  7.  Multiply  f£  by 

5.  Multiply  if  by  ft.  8.  Multiply  ff  by 

9.  What  is  the  product  of  £  into  -f  into  if  into  $  into  $  ? 

10.  What  cost  6|  yards  of  cloth,  at  4$  dollars  per  yard  ? 

Analysis.  —  4$  dollars  =f,  and  6f  yards  =-sy»-.  (Art.  197.)    Now 
^-1!^,  or  30.  (Art.  196.)     Ans.  30  dollars.     Hence, 


2  2O.  When  the  multiplier  and  multiplicand  are  both  mixed 
numbers,  they  should  be  reduced  to  improper  fractions,  and  then 
be  multiplied  according  to  the  rule  above. 

OBS.  Mixed  numbers  may  al^  be  multiplied  together,  without  reducing  them 
to  improper  fractions. 

Take,  for  instance,  the  last  example.     We  first  multiply  by  4,  Operation. 

the  whole  number.     Thus,  4  times  -f  are  -f-.  equal  to  2  and  -|  ;  **  6| 

set  down  the  |,  and  carry  the  2.     Next,  4  times  6  are  24,  and  4J 

2  to  carry  are  26.     We  then  multiply  by  J,  the  fractional  part.  "2Gf 

Thus,  |  of  6  is  3  ;  and  J  of  2  thirds  is  \.     The  sum  of  the  two  3j 

partial  products  is  30  dollars,  the  same  as  before.  30  doll*. 

11.  Multiply  6f  by  2-H-.  23.  Multiply  246-^  by  fi. 

12.  Multiply  8A  by  6f.  24.  Multiply  9  Iff-   by  -f  off. 

13.  Multiply  13|  by  17$.  25.  Multiply  1475  by  i  of  21. 
34.  Multiply  15*  by  20-f.  26.  Multiply  34i  by  i  of  68. 

15.  Multiply  30f  by  44-2\.  27.  Multiply  800  by  f  of  1000. 

16.  Multiply  63fi  by  50|.          28.  Multiply  i  of  75  by  |  of  28. 
37.  Multiply  17-ft  by  25-H-.         29.  Multiply  2-J-  by  -f  of  f  of  85. 

18.  Multiply  47M  by  l7if.  30.  Multiply  f  of  2i  by  f  of  61. 

19.  Multiply  ei^  by  32ff.  31.  Multiply  |-  of  10$  by  -f  of  8i. 

20.  Multiply  7lff  by  45-6\.  32.  Multiply  fof  16ibyf  cf  9i. 

21.  Multiply  83-^  by  61^  i  33.  Multiply  fof^o  of  20  by  25|-. 

22.  Multiply  96|f  by  72-||.  34.  Multiply  ft  of  65i  '  y  40$. 

35.  What  cost  125i  bbls.  of  flour,  at  7$  dollars  per  barrel? 

36.  What  cost  2  5  Of  acres  of  land,  at  25i  dollars  per  acre  ? 

37.  If  a  man  travels  40f  miles  per  day,  how  far  will  he  travel 
in  135$  days? 

Qv  EST.—  220.  Wheu  the  multiplier  and  multiplicand  are  mixed  numbers,  how  proceed 


130  MULTIPLICATION    OF  [SECT.  VIL 

CONTRACTIONS   IN  MULTIPLICATION  OF   FRACTIONS. 

Ex.  1.  Multiply  f  by  •§•  and  -ft-  and  i  and  f.       Up 

Operation.  Since  the  factors  3,   5   and   8  'are 

$     0      0      17      7          common  to  the  numerators  and  denom- 

_.  vx  _  NX        vx "v  — _ 

001 1     $2     22         inators,  we  may  cancel  them;    (Art. 
191 ;)  and  then  multiply  the  remain- 
ing factors  together,  as  hi  reduction  of  compound  fractions  to  sim- 
ple ones.  (Art.  199.)     Hence, 

221*  To  multiply  fractious  by  CANCELATION. 

Cancel  all  the  factors  common  both  to  the  numerators  and  de- 
nominators;  then  multiply  together  fa  factors  remaining  in  tJie 
numerators  for  a  new  numerator,  ancr  those  remaining  in  t/te  de~ 
nominators  for  a  new  denominator,  as  in  reduction  of  compound 
fractions.  (Art.  199.) 

OBS.  1.  The  reason  of  this  process  may  be  seen  from  the  fact  that  the  product 
of  the  numerators  is  divided  by  the  same  numbers  as  that  of  the  denominators, 
and  therefore  the  value  of  the  answer  is  not  altered.  ^Art.  191.) 

2.  Care  must  be  taken  that  the  factors  canceled  in  the  numerators  are  ev~ 
actly  equal  to  those  canceled  in  the  denominators. 

2.  Multiply  |  by  f  and  f.  Ans.  f . 

3.  Multiply  f  by  -I  into  £.  7.  Multiply  f  of  f  by  if. 

4.  Multiply  f  by  tV  into  £.  8.  Multiply  ft  by  if-  of  -fr- 

5.  Multiply  f  by  i  into  •£.  9.  Multiply    ri  of  4  by  -ft. 

6.  Multiply  |  by  f  of  f .  10.  Multiply   3f  by  if  of  8 

11.  Multiply  |  by  $  and  f  andf  and  f. 

12.  Multiply  i  by  -f-  and  -&  and  -fa  and  fg-. 

13.  Multiply  if  by  -f  and  -£f  and  -ft-  and  -ft  . 

L4.  Multiply  -ft  into  £f  into  f  into  •£  into  4^  into  f . 

15.  Multiply  -ft-  into  -ff  into  ||  into  ff  into  f  into  i£. 

16.  Multiply  £•£  into  -ft  into  %  into  -fe  into  -f^  into  f. 

17.  What  must  a  man  pay  for  3-^  barrels  of  flour,,  when  flour  ;» 
worth  6  dollars  a  barrel  ? 


QUSST. — 221.  How  are  fractions  multiplied  by  cancelation?  Obs.  How  does  it  appe&. 
that  this  process  will  give  the  true  answer?  What  is  necessary  to  be  observet  with  re- 
gard to  canceling  factors  1 


ARTS.  221-223.]  FRACTIONS.  131 

Analysis. — 3-J-  bbls.  is  %  of  10  Operation. 

bbls. ;    now  since  1  bbl.  costs  6  dolls.    6  price  of  1  bbl. 

dollars,  10  Ws.  will  cost  10  times  JLO^ 

as  much,  or  60  dollars.     But  we  3)60     "     of  10  bbls. 

wished  to  find  the  cost  of  only  3i  dolls. ~20      "     of  3-J-  bbls. 
barrels,   which  is  -J-  of  10  bbls. 

Therefore  if  we  take  •£  of  the  cost  of  10  bbls.,  it  will  of  cours* 
je  the  price  of  3^  bbls. 

PROOF. — 6  dolls.  X  3-^=20  dolls.,  the  same  as  before. 

Note. — In  like  manner,  when  the  multiplier  is  33  J,  333  J,  &c.,  if  we  multiply 
by  100,  1000,  &c.,  l  of  the  product  will  be  the  answer.  Hence, 

222.  To  multiply  a  whole  number  by  3|,  33|,  333£,  (fee. 
Annex  as  many  ciphers  w  the  multiplicand  as  there  are  3s  in  th6 

integral  part  of  the  multiplier  ;  then  take  •§•  of  the  number  thus 
produced,  and  the  result  will  be  the  answer  required. 

OBS.  1.  The  reason  of  this  contraction  is  evident  from  the  principle  that  an- 
nexing a  cipher  to  a  number  multiplies  it  by  10,  annexing  two  ciphers  multi- 
plies it  by  100,  &c.  (Art.  98.)  But  3$  is  £  of  10 ;  33±  is  i  of  100,  &c. ;  there- 
fore annexing  as  many  ciphers  to  th*  multiplicand,  as  there  are  3s  in  the  in- 
tegral part  of  the  multiplier,  gives  a  product  3  times  too  large  ;  consequently 
J  of  this  product  must  be  the  true  ansvier. 

2.  When  the  multiplicand  is  a  mixed  number,  and  the  multiplier  is  3£,  33 \ 
&c.,  it  is  evident  we  may  multiply  by  10,  100,  &c.,  as  the  case  may  be,  and  | 
of  the  number  thus  produced  will  be  the  answer  required. 

18.  Multiply  158  by  33^.  Ans.  5266-f. 

19.  Multiply  148  by  3i.  22.  Multiply  297  by  333|. 

20.  Multiply  256 'by  33i.  23.  Multiply  56l£  by  3£. 

21.  Multiply  1728  by  33^.  24.  Multiply  426f  by  33i. 

223.  To  multiply  a  whole  number  by  6|,  66f,  666|,  &o. 
Annex  as  many  ciphers  to  the  multiplicand  as  there  are  6s  in  the 

integral  part  of  the  multiplier ;  tJien  take  -f  of  the  number  thus 
produced,  and  the  result  will  be  the  answer  required. 

OBS.  The  reason  of  this  contraction  is  manifest  from  the  fact  that  6|  is  >  of 
10 ,  66|  of  100,  &c. 

25.  What  will  6-f  tons  of  iron  cost,  at  75  dollars  per  ton  ? 

QUEST.— 222.  How  may  a  whole  number  be  multiplied  by  3$,  33$,  &c.7  223  How 
may  a  whol  j  number  be  multiplied  by  6},  66|,  &c 


132  MULTIPLICATION    OF  [SECT.   VII. 

Analysis.   -6-|  tons  is  f  of  10  Opeiation. 

tons.     Now  if  1  ton  costs  75  dol-  dolls.  75,  price  of  1  ton 

lars,  10  tons  will  cost  10  times  as  10 

much,  or  750  dollars;  and  -f  of  3)750      "    of  10  ton& 

750  dollars,   (6-f=f  of  10,)   are  250 

500  dollars,  whnh  is  the  answer  2 

required.  dolls.   500,     "    of 6f  tons 

PROOF. — 75  dolls.  X6f =500  dolls.,  the  same  as  above. 

26.  Multiply  320  by  6f.  28.  Multiply  837  by  Gf. 

27.  Multiply  277  by  66f.  29.  Multiply  645  by  666  J. 

30.  What  will  12£  acres  of  land  cost,  at  46  dollars  per  acre? 

Analysis. — 12£  acres   is  -J-  of  lOOfc  Operation. 

«*cres ;  now  since  1  acre  costs  46  dol-  dolls.  46,  price  of  1  A* 

lars,  100  acres  will  cost  100  times  as  100 

much,  or  4600  dollars.    But  we  wished  8)4600     "  100  A. 

to  find  the  cost  of  only  1 2£  acres,  which  dolls.  575     "     12  A 
is  i  of  100  acres.     Therefore  i  of  the 
cost  of  100  acres,  will  obviously  be  the  cost  of  12£  acres. 

PROOF. — 46  dolls. Xl2i=575  dolls.,  the  same  as  before. 

Note.  In  like  manner,  if  the  multiplier  is  37J,  62 £,  or  87|,  we  may  nv  Jtip^y 
by  100,  t».nd  f,  f,  or  J  of  the  product  will  be  the  answer.  Hence, 

224    To  multiply  a  whole  number  by  12£,  37£,  62£,  or  87£. 

Annex  two  ciphers  to  the  multiplicand,  then  take  \,  -f,  -f,  or  -£, 
of  the  number  thus  produced,  as  tlie  case  may  be,  and  tlie  result 
will  be  the  answer  required. 

OBS.  The  reason  of  this  contraction  may  be  seen  from  the  fact  that  12§  is  J, 
37J  is  I,  62£  is  f ,  and  87$  is  5  of  100. 

31.  Multiply  275  by  37£.  Ans.  10312-^. 

32.  Multiply  381  by  12£.  34.  Multiply  643  by  62£. 

33.  Multiply  425  by  37£.  35.  Multiply  748  by  87£. 

225.  To  multiply  a  whole  number  by  1-f,  16|,  166|,  &c. 

Annex  as  many  ciphers  to  the  multiplicand  as  there  are  integral 
figures  in  the  multiplier.,  then  ^-  of  the  number  thus  produced  will 
be  the  product  required. 


A.RTS.  224-226. J  FRACTIONS.  133 

OBS.  The  reason  of  this  contraction  is  evident  from  the  fact  that  1|  is  -J 
of  10;  1GI  is  |  of  100;  IGGf  is  £  of  1000,  &c. 

36.  What  will  16  f  bales  of  Swiss  muslin  cost,  at  735  dollars 
per  bale  ? 

Solution. — Annexing  two  ciphers  to  735  dolls.,  it  becomes 
72500  dolls. ;  and  73500-7-6=12250  dolls.  Ans. 

37.  Multiply  767  by  If.  39    Multiply  489  by  I6f. 

38.  Multiply  245  by  16t.  40.  Multiply  563  by  166|. 

Note. — Specific  rules  might  be  added  for  multiplying  by  1^,  11-^,  lll£,  8-^, 
83j,  833J,  6$,  &c.,  but  they  will  naturally  be  suggested  to  the  inquisitive  mind 
from  the  contractions  already  given. 


DIVISION   OF   FRACTIONS. 

CASE     I. 
226.  Dividing  a  fraction  by  a  whole  number. 

Ex.  1.  If  4  yards  of  calico  cost  f  of  a  dollar,  what  will  1  yard 
cost? 

Analysis. — 1  is  1  fourth  of  4 ;  therefore  1  yard  will  cost  1 
fourth  part  as  much  as  4  yards.  And  1  fourth  of  8  ninths  of  a 
dollar,  is  2  ninths.  Ans.  f  of  a  dollar. 

Operation.  We  divide  the  numerator  of  the  fraction  by 

•§•-7-4=3  Ans.         4,  and  the  quotient  2,  placed  ovei  the  denomi- 
nator, forms  the  answer  required. 

2.  If  5  bushels  of  apples  cost  \%  of  a  dollar,  what  will  1  bushel 
cost? 

Operation.  Since  we  cannot  dh'  fle  the  numer 

H_^.5  — _ ii_         11  A  ator  ^7  t^ie  divisor  5,  without  a  re- 

12  •         12x5'  mainder,  we  multiply  the  denomina* 

tor  by  it,  which,  in  effect,  divides  the  fraction.  (Art.  188.) 

PROOF. — H  dolls.  X5=ii  dolls.,  the  same  as  above.     Henoe, 


134  DIVISION  OP  [SECT.  Vll. 

227,  To  divide  a  fraction  by  a  whole  number. 

Divide  the  numerator  by  the  whole  number,  when  it  can  be  don* 
without  a  remainder  ;  but  when  this  cannot  be  done,  multiply  the 
denominator  by  the  whole  number. 

3.  What  is  the  quotient  of  ^g-  divided  by  5  ? 


First  Method.  Second  Method. 

15  3  15  15         15  3 

^5  =      Ans'  +5 


4.  Divide  if  by  9.  7.  Divide  -fi  by  12. 

5.  Divide  ff  by  7.  8.  Divide  W  by  25. 

6.  Divide  ffr.  by  16.  9.  Divide  -fJHJ-  by  29. 

CASE    II. 
228*  Dividing  a  fraction  by  a  fraction. 

10.  At  -J-  of  a  dollar  a  basket,  how  many  baskets  of  peaches 
can  you  buy  for  £  of  a  dollar  ? 

Analysis.  —  Since  f  of  a  dollar  will  buy  1  basket,  £  of  a  dollar 
will  buy  as  many  baskets  as  £  is  contained  times  in  £  ;  and  %  is 
contained  in  £,  4  times.  Ans.  4  baskets. 

11.  At  f  of  a  dollar  per  yard,  how  many  yards  of  cloth  can  be 
bought  for  f  of  a  dollar  ? 

OBS.  1.  Reasoning  as  before,  5  of  a  dollar  will  buy  as  many  yards,  as  |  of 
a  dollar  is  contained  in  J.  But  since  the  fractions  have  different  denomina- 
tors, it  is  plain  we  cannot  divide  one  numerator  by  the  other,  as  we  did  in  the 
last  example.  This  difficulty  may  be  remedied  by  reducing  the  fractions  to  a 
common  denominator.  (Art.  200.) 

First  Operation. 

•f  and  f  reduced  to  a  common  denominator,  become  ^  and  £f  . 
(Art.  200.)  Now  fi-j-if  =-  H  I  and  -?i=l-ft-.  ^™.  1-fr  yards. 

OBS.  2.  It  will  be  perceived  that  no  use  is  made  of  the  common  denominator^ 
after  it  is  obtained.  If,  therefore,  we  invert  the  divisor,  and  then  multiply  the 
two  fractions  together,  we  shall  have  the  same  result  as  before. 

Second  Operation. 
(divisor  in  verted)  =f£,  or  IT\  yards,  the  same  as 


QUEST  —227.  How  is  a  fraction  divided  by  a  whole  number  *? 


ARTS.  227-230  J  FRACTIONS.  J35 

229.  Hence,  to  divide  a  fraction  by  a  fraction. 

I.  If  the  given  fractions  have  a  common  denominator.,  divide  tht 
numerator  of  the  dividend  by  the  numerator  of  the  divisor. 

II.  When  the  fractions  have  not  a  common  denominator,  invert 
the  divisor,  and  proceed  as  in  multiplication  of  fractions.  (Art.  219.) 

OBS.  1.  When  two  fractions  have  a  common  denominator,  it  is  plain  one 
nuwrator  can  be  divided  by  the  other,  as  well  as  one  whole  number  by  an- 
sike" ;  for,  the  parts  of  the  two  fractions  are  of  the  same  denomination. 

2  When  the  fractions  do  not  have  a  common  denominator,  the  reason  that 
inverting  the  divisor  and  proceeding  as  in  multiplication,  will  produce  the  triti 
Answer,  is  because  this  process,  in  effect,  reduces  the  two  fractions  to  a  com- 
mon denominator,  and  then  the  numerator  of  the  dividend  is  divided  by  the 
numerator  of  the  divisor.  Thus,  reducing  the  two  fractions  to  a  common  de- 
nominator, we  multiply  the  numerator  of  the  dividend  by  the  denominator  of 
the  divisor,  and  the  numerator  of  the  divisor  by  the  denominator  of  the  divi- 
dend ;  (Art.  '200  ;)  and,  then  dividing  the  former  product  by  the  latter,  we  have 
ihe  same  combination  of  the  same  numbers  as  in  the  rule  above,  which  will  con- 
sequently produce  the  same  result. 

We  do  not  multiply  the  two  denominators  together  for  a  common  denomina- 
tor ;  for,  in  dividing,  no  use  is  made  of  a  common  denominator  when  found, 
therefore  it  is  unnecessary  to  obtain  it.  (Art.  228.  Obs.  2.) 

The  object  of  inverting  the  divisor  is  simply  for  convenience  in  multiplying. 

3.  Compound  fractions  occurring  in  the  divisor  or  dividend,  must  be  re- 
duced to  simple  ones,  and  mixed  numbers  to  improper  fractions. 

230.  The  principle  of  dividing  a  fraction  by  a  fraction  may 
also  be  illustrated  in  the  following  manner.     Thus,  in  the  last 
example, 

Dividing-  the  dividend  •£  by  2,  the  quo-  Operation: 

tient  is  -ft.  (Art.  188.)     But  it  is  required  -£-^2  =  -ft 

to  divide  it  by  1  third  of  two  ;  consequently          &X%=H 
the   -ft-  is  3  times  too  small  for  the  true       And  -^=1-^-  Ans* 
quotient ;    therefore  multiplying  -ft  by  3, 
will  give  the  quotient  required  ;  and  -ft-X3=-fi,  or  1-^-. 

Note. — By  examination  the  learner  will  perceive  that  this  process  is  precisely 

QTEST. — 229.  How  is  one  fraction  divided  by  another  when  they  have  a  common  de 
nominator  1  How,  when  they  have  not  common  denominators  ?  Oba.  When  the  fractions 
have  a  common  denominator,  how  does  it  appear  that  dividing  any  numerator  by  the  other 
will  give  the  true  answer  1  When  the  fractions  have  not  a  common  denominator,  how 
does  it  appear  that  inverting  the  divisor  and  proceeding  as  in  multiplication  will  give  the 
true  answer  1  What  is  the  object  of  inverting  the  divisor  7  How  proceed  when  the  divisor 
or  dividend  are  compound  fractions  or  mixed  numbers  7 


136  DIVISION  OF  [SECT.  VII 

the  same  in  effect  as  the  preceding;  for,  in  both  cases  the  denominator  of  thfc 
dividend  is  multiplied  by  the  numerator  of  the  divisor,  and  the  numerator  of 
the  dividend,  by  the  denominator  of  the  divisor. 

12.  Divide  f  of  J  by  2£.  Ans.  if,  or  ft. 

13.  Divide  8-f  by  3£.  Ans.  -|f,  or  2££. 

14.  Divide  £?  by  ff-.  16.  Divide  55£  by  16f. 

15.  Divide  ff  by  if.  17.  Divide  461  by  68f. 

231*  The  process  of  dividing  fractions  may  often  be  con- 
tf  acted  b*  canceling  equal  factors  in  the  divisor  and  dividend ; 
(Art.  146  ;)  or,  after  the  divisor  is  inverted,  by  canceling  factors 
which  are  common  to  the  numerators  and  denominators.  (Art.  191.) 

18.  Divide  \  of  *  of  -ft  by  £  of  -f  of  f. 

Operation.  For  convenience  we  arrange  the  numera- 

1  tors,  (which   answer  to  dividends,)  on  the 

right  of  a  perpendicular  line,  and  the  de- 
ll $  nominators,  (which  answer  to  divisors,)  on 
the  left ;  then  canceling  the  factors,  2,  3,  4, 
and   7,   which  are  common  to  both   sides, 
(Art.  151,)  we  multiply  the  remaining  fac- 
lYJ5=A-  Ans         ^ors  *n  ^he  numerators  together,  and  those 
remaining  in  the   denominators,    as  in  the 
rule  above.     Hence, 

232.  To  divide  fractions  by  CANCELATION. 

Having  inverted  the  divisor,  cancel  all  the  factors  common  both 
to  the  numerators  and  denominators,  and  the  product  of  those  re- 
maining on  the  right  of  the  line  placed  over  tlieprod.net  of  those 
remaining  on  the  left,  will  be  the  answer  required. 

OBS.  1.  Before  arranging  the  terms  of  the  divisor  for  cancelation,  it  is  always 
necessary  to  invert  them,  or  suppose  them  to  be  inverted. 

2.  The  reason  of  this  contraction  is  evident  from  the  principle,  that  if  the 
numerator  and  denominator  of  a  fraction  are  both  divided  by  the  same  w  ^Jf^rt 
the  value  of  the  fraction  is  not  altered.  (Arts.  148,  191.) 

19.  Divide  18f  by  6-f.  Answer  3. 

QI:ESI. — 232.  How  divide  fractions  by  cancelation?  How  arrange  the  terms  of  fh« 
given  fractions  "t  Obs.  What  must  be  done  to  the  divisor  before  arranging  its  terms  ?  Ho* 
does  it  appear  that  this  contraction  will  give  the  true  answer  1 


ARTS.  231-234.]  FRACTIONS.  137 

20.  Divide  £  of  -f  by  f  of  -ft-      23.  Divide  f  of  7f-  by  f  of  f . 

21.  Divide  f  of  ft  by  6f.  24.  Divide  f  of  f  of  -f  by  f 

22.  Divide  15|  by  -ft-  of  -f.          25.   Divide  f-f  of  7  by  fi  of  42. 
26.  Divide  H  of  li  of  A  of  |i  by  -&  of  ff  of  f  of  5. 


CASE    III. 

233*  Dividing  a  whole  number' by  a  fraction. 

27.  How  many  pounds  of  tea,  at  -^  of  a  dollar  a  pound,  sari  be 
oouglit  for  15  dollars  ? 

Analysis. — Since  f  of  a  dollar  will  buy  1  pound,  15  dollars  will 
buy  as  many  pounds  as  f-  is  contained  times  in  15.  Reducing  the 
dividend  15,  to  the  form  of  a  fraction,  it  becomes  -4s-;  (Art.  197. 
Obs.  1 ;)  then  inverting  the  divisor  and  proceeding  as  before,  we 
have  -"^Xi^V-,  or  20.  Ans.  20  pounds. 

Or,  we  may  reason  thus :  -|  is  contained  in  15,  as  many  times 
as  there  are  fourths  in  15,  viz:  60  times.  But  3  fourths  will  be 
contained  in  15,  only  a  third  as  many  times  as  1  fourth,  and 
60-r-3  =  20,  the  same  result  as  before.  Hence, 

234*  To  divide  a  whole  number  by  a  fraction. 

Reduce  the  whole  number  to  the  form  of  a  fraction,  (Art.  197. 
Obs.  1,)  and  then  proceed  according  to  the  rule  for  dividing  a 
fraction  by  a  fraction.  (Art.  229.) 

Or,  multiply  the  whole  number  by  the  denominator,  and  divide 
the  product  by  the  numerator. 

OBS.  1.  When  the  divisor  is  a  mixed  number,  it  must  he  reduced  to  an  im- 
proper fraction  ;  then  proceed  as  above. 

Or,  reducing  the  dividend  to  a  fraction  having  the  same  denominate,  (Art. 
197.  Obs.  2,)  we  may  divide  one  numerator  by  the  other.  (Art.  229.  I.) 

2.  If  the  divisor  is  a  unit  or  1,  the  quotient  is  equal  to  the  dividend  ;  if  the 
divisor  is  greater  than  a  unit,  the  quotient  is  less  than  the  dividend ;  and  if  the 
divisor  is  less  than  a  unit,  the  quotient  is  greater  than  the  dividend. 

28.  How  much  cloth,  at  3£  dollars  per  yard,  can  you  buy  for 
28  dollars  ? 

QUEST.— 234.  How  is  a  whole  number  divided  by  a  fraction  ?  Obs  How  by  a  mixed 
aumber  1 


'38  DIVISION    OF  L^ECT' 

Operation.  Since  the  divisor  is  a  mixed  number, 

3^)28  we  reduce  it  to  halves  ;  we  also  reduce 

the  dividend  to  the  same  denominator ; 

7)  56  halves.  (Art.  197.  Obs.  2  ;)  then  divide  one  nu> 

Ans.  8  yards.  merator  by  the  other.  (Art.  229.  I.) 

29.  Divide  75  by  f.  32.  Divide  145  by  12|. 

30  Divide  96  by  f.  33.  Divide  237  by  25f. 

31  Divide  120  by  10^.  34.  Divide  425  by  31$. 

CONTRACTIONS   IN   DIVISION   OF   FRACTIONS. 

235.  When  the  divisor  is  3|,  33i,  333^,  &c. 

Multiply  the  dividend  by  3,  divide  the  product  by  10,  100,  or 
1000,  as  tlie  case  may  be,  and  the  result  will  be  the  true  quotient. 
(Art.  131.) 

OBS.  The  reason  of  this  contraction  will  be  understood  from  the  principle, 
that  if  the  divisor  and  dividend  are  both  multiplied  by  the  same  number,  the 
quotient  will  not  be  altered.  (Art.  14G.)  Thus  3^X3=10;  33JX3=100; 
333*X3=1000,  &c. 

35.  At  3-^  dollars  per  yard,  how  many  yards  of  cloth  can  be 
bought  for  561  dollars9 

Operation.  We    first  multiply   the  dividend   by  3, 

dolls.  561  then  divide  the  product  by  10;  for,  mul- 

3  tiplymg  the  divisor  3-^  by  3,  it  becomes  10. 

1|0)1C8|3  (Art.  146.) 
Ans.  168/0  yds. 

36.  Divide  687  by  33£.  Ans.  20-fljV. 

37.  Divide  453  by  33^,  38.  Divide  2783  by  333£. 

236.  When  the  divisor  is  If,  16|,  166|,  &c. 

Multiply  the  dividend  by  6,  and  divide  the  product  by  10,  100, 
or  1000,  as  tJie  case  may  be. 

OBS.  This  contraction  also  depends  upon  the  principle,  that  if  the  divisor 
and  dividend  are  both  multiplied  by  the  same  number,  the  quotient  will  riat  b« 
altered.  (Art.  146.)  Thus,  lfX6=lO;  16|X6=100;  166§X6=1000,  &c. 


ARTS.  235-239.]  FRACTIONS.  J39 

39.  What  is  the  quotient  of  725  divided  by  16|  ? 
Solution.— 725X6—  4350;  and  4350-rlOO  =  43i     Ans. 

40.  Divide  367  by  If.  42.  Divide  849  by  16-f. 

41.  Divide  507  by  16-f.  43.  Divide  1124  by  166-f. 

237.  VThen  the  divisor  is  H,  1H,  11H,  &c. 

Multiply  the  dividend  by  9,  and  divide  the  product  by  10,  100, 
*r  1000,  as  the  case  may  be. 

OBS.  This  contraction  depends  upon  the  same  principle  as  the  preceding 
Thus,  1|X9=10;  11|X9=100;  HlfX9=lOOO,  &c. 

44.  Divide  587  by  Hi. 

Solution.— 587X9  =  5283,  and  5283-7-100  =  52^    Ans. 

45.  Divide  861  by  H.  Ans.  774-rV. 

46.  Divide  4263  by  Hi.  47.  Divide  6037  by  llli- 

Tvotc. — Other  methods  of  contraction  might  be  added,  but  they  will  rjaturally 
suggest  themselves  to  the  student,  as  he  becomes  familiar  with  the  principles 
of  fractions. 

238*  From  the  definition  of  complex  fractions,  and  the  man- 
ner of  expressing  them,  it  will  be  seen  that  they  arise  from  di- 

AL 

vision  of  fractions.  (Art.  183.)  Thus,  the  complex  fraction  y-,  is 
the  same  as  -f-7-f- ;  for,  the  numerator,  4£=f ,  and  the  denomina- 
tor 1-J-^-f- ;  but  the  numerator  of  a  fraction  is  a  dividend,  and  the 
denominator  a  divisor.  (Art.  184.)  Now,  -|-7-J=li-.  which  is  a 
simple  fraction.  Hence, 

239*  To  reduce  a  complex  fraction  to  a  simple,  one. 
Consider  the  denominator  as  a  divisor,  and  proceed  as  in  dims- 
ion  of  fractions.  (Arts.  229,  232.) 

OBS.  The  reason  of  this  rule  is  evident  from  the  fact  that  the  denominator 
of  a  fraction  denotes  a  divisor,  and  the  numerator,  a  dividend;  (Art.  184; 
nence  the  process  required,  is  simply  performing  the  division  which  is  ex- 
pressed by  the  given  fraction. 

QUEST. — 238  P  om  what  do  complex  fractions  arise  ?  239.  How  reduce  them  to  sii»- 
plc  fractions  7 

T.H.  t 


140  COMPLEX  {SECT.   VII 


48.  Reduce  -|  to  a  simple  fraction. 
Solutim.—  4|=-KS  and  7}=^.  (Art.  197.) 

NOW  Y-r-^YX  -gV,  or  -ff      ^W5. 

Reduce  the  following  complex  fractions  to  simple  o  nes  : 

8  5-£ 

49.  Reduce—.  53.  Reduce-^. 

34  T6 

51  7 

50.  Reduce  y.  54.  Reduce-^. 

2-2  -2-5- 

51.  Reduce  -|.  55.  Reduce—. 

07  90 

6i  44 

52.  Reduce      .  56.  Reduce      . 


24O.  To  multiply  compbx  fractions  together. 

First  reduce  the  complex  fractions  to  simple  ones  ;  (Art.  £39;) 
then  arrange  the  terms,  and  cancel  the  common  factors,  as  in  mul- 
tiplication of  simple  fractions.  (Art.  219.) 

OBS.  The  terms  of  the  complex  fractions  may  be  arranged  for  reducing  them 
to  simple  o/ies,  and  for  multiplication  at  the  same  time. 

57.  Multiply  g  by  g. 

Operation.  The  numerator  3^=-|.    (Art.  197.)     Place 

the  7  on  the  right  hand  and  2  on  the  left  of 
the  perpendicular  line.  The  denominator  2-f 
=Jsa,  which  must  be  inverted;  (Art.  239  ;) 
i.  e.  place  the  12  on  the  right  and  the  5  on 
the  left  of  the  line,  1-f  =-V,  and  4-£=f,  both 
of  wVich  must  be  arranged  in  the  same  man- 
ner as  the  terms  of  the  multiplicand.  Now,  canceling  the  com- 
mon factors,  we  divide  the  product  of  those  remaining  on  the  right 
of  the  line  by  the  product  of  those  on  the  left,  and  the  answer 
is  |.  (Art.  219.) 


QUKST.— 240.  How  are  complex  fractions  multiplied  together  ?    241.  How  is  one  com- 
plex fraction  divided  by  another  ? 


ARTS.  240-242.  J  FRACTIONS.  141 

58.  Multiply  ?±  by  ~  60.  Multiply  |  by  |  into  ~. 

^4  ^3  4  T  f 

59.  Multiply  i|  by  ^  .  61.  Multiply  ^  by  ?|  into  j|. 

241*  To  divide  one  complex  fraction  by  another. 

Reduce  the  complex  fractions  to  simple  ones,  then  proceed  as  in 
'{vision,  of  simple  fractions.  (Arts.  229,  239.) 

62.  Divide   §byj. 

4i     9     4     36        ,+      144 
*-5f=5Xr-ig,  and  I2=3X?=2-I.  (Art.  239.) 

36      4      30     21      756 


9X4  1X4 

Or,  since  the  given  dividend  =  -  and  the  divisor  =  -  r 

9X4     3  ^  Y 
then  —  —  X  YTH  ~  the  ans  wer-  (  Art-  2  3  1  •) 


9X4     3XV     0X^X3XV     21 
But,  (Art.  232.)    —  x  —  =  =>  or  10-  An* 


63.  Divide  |  by  |.  64.  Divide  ^|  by  ?|. 


APPLICATION    OF    FRACTIONS. 

Ex.  1.  A  merchant  bought  15-£  yards  of  domestic  flan- 
nel of  one  customer,  19^  of  another,  12-£  of  another,  and  41-$^-  of 
another :  how  many  yards  did  he  buy  of  all  ? 

2.  A  grocer  sold  16-J-  Ibs.  of  sugar  to  one  customer,  112£  to 
another,  and  33^-  to  another  :  how  many  pounds  did  he  sell  ? 

3.  A  clerk  spent  26f  dollars  for  a  coat,  9f  dollars  for  pants, 
6f  dollars  for  a  vest,  .5^-  dollars  for  a  hat,  and  6-J-  dollars  for  a 
pair  of  boots  :  how  much  did  his  suit  cost  him  ? 

4.  A  man  having  bought  a  bill  of  goods  amounting  to  85-^g-  dol- 
lars, handed  the  clerk  a  bank  note  of  100  dollars:  how  much 
change  ought  he  *a  receive  back? 


142  DIVISION  OF  [SECT.  VII. 

5.  A  lady  went  a  shopping  with   135}-  dollars  in  her  purse* 
she  paid  17^  dollars  for  silk,  3f  dollars  for  trimmings,  37£  dol- 
lars for  a  shawl,  and  14f  dollars  foi  a  muff:  how  much  money 
had  she  left  ? 

6.  A  man  having  1563^-  dollars,  spent  365f  dollars,  and  lost 
fif>2£  dollars  :   how  much  had  he  left  ? 

7.  What  will  563  sLeep  cost,  at  2f  dollars  per  head  ? 

8.  What  cost  748  barrels  of  flour,  at  7-f  dollars  per  barrel  ? 

9.  What  cost  378f  yards  of  cloth,  at  4  dollars  per  yard? 

10.  What  cost  1121-^-  Ibs.  of  tea,  at  5  shillings  per  pound? 

11.  What  cost  430  gallons  of  oil,  at  H  dollar  per  gallon? 

12.  What  cost  ff  of  an  acre  of  land,  at  150  dollars  per  acre  ? 

13.  A  man  worth  25000  dollars,  lost  H  of  it  by  fire :  what 
was  the  amount  of  his  loss  ? 

14.  A  garrison  had  856485  pounds  of  flour ;  after  being  block- 
aded 60  days,  it  was  found  that  £$£•  of  it  were  consumed :  how 
many  pounds  of  flour  were  left  ? 

15.  At  I7i  dollars  per  ton,  what  cost  103-J-  tons  of  hay? 

16.  How  many  bushels  of  corn  will  115f  acres  produce,  at  31-} 
bushels  per  acre  ? 

17.  What  cost  675-|-  tons  of  iron,  at  45f  dollars  per  ton? 

18.  If  a  ship  sails  140-fV  miles  per  day,  how  far  will  she  sail 
in  49 £  days? 

19.  If  a  Railroad  car  should  run  4l£  miles  per  hour,  how  far 
would  it  go  in  12  days,  running  10£  hours  per  day? 

20.  A  young  man  having  a  patrimony  of  12234  dollars,  spent 
~  of  it  in  dissipation  :  how  much  had  he  left  ? 

21.  At  -f-  of  a  dollar  per  yard,  how  many  yards  of  satinet  can 
be  bought  for  124  dollars? 

22.  How  many  pounds  of  tea,  at  f  of  a  dollar  a  pound  can  you 
buy  for  331  dollars? 

23.  How  many  gallons  of  molasses,  at  |  of  a  dollar  per  gallon 
can  you  buy  for  235  dollars  ?^ 

24.  At  8  pence  a  pound,  how  many  pounds  of  sugar  can  you 
buy  for  163-J-  pence? 

25.  At  5-J-  pen;  e  a  yard,  how  many  yards  of  lace  can  be  bought 
for  279  pence  ? 


ART.  242.]  FRACTION?.  143 

26.  A  dairy-man  has  229£  pounds  of  butter  wlich  he  wishes 
to  pack  in  boxes  containing  8%  pounds  each :  how  many  boxes 
will  it  require  ? 

27.  A  farmer  wishes  to  put  384  bushels  of  apples  into  barrels, 
each  containing  2£  bushels  :  how  many  barrels  will  it  require  ? 

28.  If  4 1  yards  of  cloth  make  a  suit  of  clothes,  how  many  suits 
will  141£  yards  make  ? 

29.  One  rod  contains  5-£  yards :  how  many  rods  are  there  in 
210  yards? 

30.  A  merchant  paid  204^  dollars  for  57  yards  of  cloth :  how 
much  was  that  per  yard  ? 

31.  A  grocer  sold  50  barrels  of  flour  for  31 1£  dollars:  what 
did  he  get  per  barrel  ? 

32.  A  merchant  wishes  to  lay  out  657£  dollars  for  wheat,  which 
is  worth  !•£  of  a  dollar  a  bushel :  how  much  can  he  buy  ? 

33.  At  18-f  cents  a  dozen,  how  many  dozen  of  eggs  can  you 
buy  for  87£  cents  ? 

34.  A  grocer  sold  15£  pounds  of  coffee  for  93f  cents:  how 
much  was  that  a  pound  ? 

35.  A  shopkeeper  sold  16£  yards  of  satin  for  163-fr  shillings: 
how  much  was  that  per  yard  ? 

36.  Bought  19  sacks  of  wool  for  250f  dollars:  what  was  that 
per  sack  ? 

37.  Paid  575f  dollars  for  96$  yards  of  cloth:  what  was  the 
cost  per  yard  ? 

38.  Paid  1565^  dollars  for  iron,  valued  at  3 7-J-  dollars  per  ton: 
how  many  tons  were  bought  ? 

39.  Paid  1315^  dollars  for  the  transportation  of  1286  barrels 
of  pork :  what  was  that  per  barrel  ? 

40.  Bought  37oi  pounds  of  indigo  for  65  2|  dollars :  what  was 
the  cost  per  pound  ? 

41.  Paid  1679i  dollars  for  475  kegs  of  lard:  how  much  was 
that  per  keg  ? 

4.2.  If  an  army  consumes  563^  pounds  of  meat  per  day,  how 
long  will  150000  pounds  supply  it? 

43.  The  cost  of  making  25i  miles  of  Railroad  was  856?35|  dol« 
lars :  what  was  the  cost  per  mile  ? 


144  COMPOUND    NUMBERS.  [SECT.  VIII. 

SECTION    VIII. 
COMPOUND    NUMBERS. 

ART.  243*  Numbers  which  express  things  of  tl  e  same  kind 
or  denomination,  are  called  Simple  Numbers.  Thus,  3  oranges, 
7  books,  12  chairs,  &c.,  are  simple  numbers. 

Numbers  which  express  things  of  different  kinds  or  denomina- 
tions, as  the  divisions  of  money,  weight,  and  measure,  are  called 
COMPOUND  NUMBERS.  Thus,  15  shillings  6  pence;  10  bushels 
3  pecks,  &c.,  are  compound  numbers. 

OBS.  The  origin  of  Compound  Numbers  is  ascribed  to  the  wants  and  neces- 
sities of  the  earlier  ages  of  the  world.  Their  divisions  and  subdivisions  are 
generally  irregular,  and  seem  to  have  been  suggested  by  the  caprice,  or  the  lim- 
ited business  transactions  of  the  rude  ages  of  antiquity.  It  is  much  to  be  re- 
gretted, both  on  account  of  simplicity  and  their  adaptation  to  scientific  pur- 
poses, that  their  different  denominations  were  not  graduated  according  to  the 
law  of  increase  in  the  decimal  notation. 

Note. — Compound  Numbers,  by  some  authors,  are  called  Denominate 
Numbers. 

FEDERAL   MONEY. 

214.  federal  Money  is  the  currency  of  the  United  States. 
The  denominations  are,  Eagles,  Dollars,  Dimes,  Cents,  and  Mills. 

10  mills  (m.)  make  1  cent,  marked  ct. 

10  cents  "     1  dime,                     "  d. 

10  dimes  "     1  dollar,                   "  doll,  or  $. 

10  dollars  «     1  eagle,                    «  E. 

OBS.  1.  Federal  money  was  established  by  Congress,  Aug.  8,  1786.  It  rt 
based  upon  the  principles  of  the  decimal  notation.  The  law  of  increase  OL 
radix,  is  the  same  as  that  of  simple  numbers,  and  it  is  confessedly  one  of  th« 
most  simple  and  comprehensive  systems  of  currency  in  the  civilized  world.  Pre- 
vious to  its  adoption,  English  or  sterling  money  was  the  principal  currency  of 
the  country. 

Q.UEST. — 343.  What  are  simple  numbers  ?  What  are  compound  numbers  ?  244.  Wha 
fc  Federal  money  ?  Recite  the  table.  Obs.  When  and  by  whom  was  it  established  I 


ARTS,  243-246.  J       COMPOUND  NUMBERS.  145 

2.  The  names  of  the  coins  or  denominations  less  than  a  dollar,  are  signifi- 
cant of  their  value.     The  term  dime,  is  derived  from  the  French  disme,  which 
signifies  ten;  the  terms  cent  and  mill,  are  from  the  Latin  centum  and  mille. 
the  former  of  which  signifies  a  hundred,  and  the  latter  a  tkoitsand.     Thus, 
10  dimes*,  100  cents,  or  1000  mills,  make  1  dollar. 

3.  The  sign  ($),  which  is  prefixed  to  Federal  money,  is  called  the  Dollar 
mark.     It  is  said  to  be  a  ccntraction  of  "  U.  S.,"  the  initials  of  United  Slates^ 
which  were  originally  prefixed  to  sums  of  money  expressed  in  the  Federal 
currency.     At  length  the  two  letters  were  moulded  or  merged  into  a  single  char- 
acter by  dropping  the  curve  of  the  U,  and  writing  the  S  over  it.     Thus,  th« 
•um  of  seventy-five  dollars,  which  was  originally  written  "  U.  S.  75  dollars, 
is  now  written  $75. 

245*  The  national  coins  of  the  United  States  are  of  three 
kinds,  viz  :  gold,  silver,  and  copper. 

1.  The  golds  coin  are  the  eagle,  the  double  eagle*  half  eagle, 
quarter  eagle,  and  gold  dollar* 

The  eagle  contains  258  grains  of  standard  gold  ;  the  half  eagle 
and  quarter  eagle  like  proportions.! 

2.  The  silver  coins  are  the  dollar,  half  dollar,  quarter  dollar, 
the  dime,  half  dime,  and  three-cent-piece. 

The  dollar  contains  412£  grains  of  standard  silver;  the  others 
like  proportions.]- 

3.  The  copper  coins  are  the  cent,  and  half  cent. 

The  cent  contains  1C8  grains  of  pure  copper;  the  half  cent,  a 
like  proportion.^  Mills  are  not  coined. 

OBS.  The  fineness  of  gold  used  tor  coin,  jewelry,  and  other  purposes,  also 
the  gold  of  commerce,  is  estimated  by  the  number  of  parts  of  gold  which  it 
contains  Pure  gold  is  commonly  supposed  to  be  divided  into  24  equal  parts, 
"•alied  carats.  Hence,  if  it  contains  10  parts  of  alloy,  or  some  baser  metal,  it  is 
«aid  to  be  14  carats  fine  ;  if  5  parts  of  alloy,  19  carats  fine  ;  and  when  abso- 
lutely pure,  it  is  24  carats  fine. 

246.  The  present  standard  for  both  gold  and  silver  coin  o. 
the  United  States,  by  Act  of  Congress,  1837,  is  900  parts  cf  pur* 


Of  how  many  kinds  are  the  coins  of  the  United  States  ?  What  are  they 
What  are  the  gold  coins  1  The  silver  coins  ?  The  copper?  Obs.  How  is  the  fineness  of 
g«>ld  estimated  ?  Into  how  many  carats  is  pure  gold  supposed  to  be  divided  1  When  i 
contains  10  parts  of  alloy,  how  fine  is  it  said  to  be  ?  5  parts  of  alloy?  246.  What  is  the 
present  standard  for  the  gold  am*  silver  coin  of  tho  United  States  ?  What  is  the  alloy  of 
fold  com  t  What  of  silver  coin  ? 

*  Added  by  Act  of  Congress,  1849.  f  According  to  Act  of  Congress,  1837. 


146  COMPOUND    NUMBERS.  [SECT.  VIII 

metal  by  weight  to  100  parts  of  alloy.  The  alloy  of  gold  coin  is 
composed  of  silver  and  copper,  the  silver  not  to  exceed  the  cop- 
per in  weight.  The  alloy  of  silver  coin  is  pure  copper. 

Note. — The  original  standard  for  the  gold  coin  of  the  United  States  by  Act 
of  Congress,  1792,  was  22  parts  of  pure  gold  to  2  parts  of  alloy ;  the  alloy 
consisting  of  1  part  silver  and  1  part  copper. 

The  original  standard  for  the  silver  coin  was  1489  parts  of  pure  silver  to  1 79 
parts  of  alloy;  the  alloy  being  of  pure  copper. 

The  tagle  by  the  same  act  contained  270  grains  of  standard  gold.  The  djl~ 
ir  contained  416  grains  of  standard  silver.  The  cent  contained  11  penny- 
weights, or  264  grains  of  pure  copper. 

STERLING  MONEY. 

247*  English  or  Sterling  Money  is  the  national  currency  of 
Great  Britain. 

4  farthings  (qr.  or/ar.)  make  1  penny,      marked  d. 

12  pence  "     1  shilling,          "  s. 

20  shillings  "     1  pound,  or  sovereign,  £. 

21  shillings  "     1  guinea. 

OBS.  1.  It  is  customary,  at  the  present  day,  to  express  farthings  in  fractions 
of  a  penny.  Thus,  1  qr.  is  written  i  d. ;  2  qrs.  J  d. ;  3  qrs.  f  d. 

2.  The  Pound  Sterling  is  represented  by  a  gold  coin,  called  a  Sovereign. 
According  to  Act  of  Congress,  1842,  its  value  is  4  dollars  and  84  cents.     Hence, 
the  value  of  a  shilling  is  24-£  cents  ;  that  of  a  penny  2  cents,  very  nearly. 

3.  The  letters  £.  s.  d.  and  q.  are  the  initials  of  the  Latin  words,  libra,  soli- 
dus,  denarius,  and  quadrans,  which  respectively  signifiy  a  pound,  shilling, 
penny,  a.ndfartki7ig  or  quarter.     The  mark  /,  which  is  often  placed  between 
shillings  and  pence,  is  a  corruption  of  the  longy. 

Note. — 1.  Sterling  money  is  supposed  by  some  to  have  received  its  name 
from  the  Easterlings,  who  it  is  said  first  coined  it ;  others  think  it  is  so  called 
to  distinguish  it  from  stocks,  &c.,  whose  value  is  nominal. 

2.  The  pound  is  so  called,  because  in  ancient  times  the  silver  for  it  weighed 
a  pound  Troy.  A  pound  Troy  of  silver  is  now  worth  66  shillings,  or  j£3,  6s. 

The  Guinea  is  so  called,  because  the  gold  of  which  it  was  originally  made, 
was  brought  from  Guinea,  on  the  coast  of  Africa. 

248.  The  following  denominations  are  frequently  met  with, 
viz:  the  Groat = 4d. ;  the  Crown  =  5s. ;  the  Nobler 65.  Sd. ; 

QUEST.— 247.  What  is  Sterling  Money!  Repeat  the  Tal  le  ?  Obs.  Hew  are  farthings 
usually  expressed  1  How  is  a  pound  sterling  represented  ?  What  is  its  value  in  dWait 
and  cents  1 


ARTS.  247-250.]        COMPOUND  NUMBERS.  147 

the  Angel=10s.;  the  Mark=13s.  4d. ;  the  Pistole=lGs.  IQd. 

the  Moidore— 27s. 

OBS.  The  present  standard  gold  coin  of  Great  Britain,  consists  of  22  part* 
pure  gold,  and  2  parts  of  copper.* 

The  weight  of  a  Sovereign  or  £,  is  5  pwts.,  3^-^  grains. 

The  standard  silver  coin  consists  of  37  parts  of  pure  silver,  and  3  parts  of 
copper.  The  weight  of  a  shilling  is  3  pwts.  15^f  grs. 

In  copper  coin  24  pence  weigh  1  pound  avoirdupois. 

TROY   WEIGHT. 

249.  Troy  Weight  is  used  in  weighing  gold,  silver,  jewels, 
liquors,  &c.,  and  is  generally  adopted  in  philosophical  experiments. 

24  grains  (gr.)         make  1  pennyweight,          marked   pwt. 
20  pennyweights          "     1  ounce  "          oz. 

12  ounces  "     1  pound,  «          Ib. 

OBS.  1.  The  abbreviation  oz.,  is  derived  from  the  Spanish  onza,  which  sig- 
nifies an  ounce. 

2.  The  standard  of  Weights  and  Measures  is  different  in  different  countries, 
and  in  different  States  of  the  Union.  In  1834,  the  Government  of  the  United 
States  adopted  a  uniform  standard,  for  the  use  of  the  several  Custom-houses 
and  other  purposes. 

250.  The  standard  Unit  of  Weight  adopted  by  the  Govern- 
ment, is  the    Troy   Pound  of   the  United   States  Mint.     It  is 
equal  to  22.7^4422  cubic  inches  of  distilled  water,  at  its  max- 
imum   density,f  the    barometer  standing  at    30    inches,  and   is 
identical  with  the  Imperial  Troy  pound  of  Great  Britain,  estab- 
lished by  Act  of  Parliament,  in  1826.J 

OBS.  The  weights  and  measures  in  present  use,  were  derived  from  very  im~ 
'perfect  and  variable  standards.  A  grain  of  wheat,  taken  from  the  middle  of 
the  ear  or  head,  and  being  thoroughly  dried,  was  the  original  element  of  aL 
weights  used  in  England,  and  was  thence  called  a  grain.  At  first,  a  weight 

QUEST.— 249.  In  what  is  Troy  Weight  used  1  Repeat  the  Table  1  Ola.  Do  all  the 
States  have  the  same  standard  of  weights  and  measures  1  250.  What  is  the  standard 
•init  of  weight  adopted  by  the  Government  of  the  United  States  ?  JVbfe.  When  was  Troy 
Weight  introduced  into  Europe  ?  From  what  was  its  name  derived  1 

*  Hind's  Arithmetic  ;  also,  Hutton's  Mathematics. 

t  The  maximum  density  of  water,  according  to  Mr.  Hassler,  is  at  the  temperature  ot 
39-83  deg.  Fahrenheit. 

t  The  Troy  pound  of  the  U.  S.  Mint,  is  an  exact  copy,  by  Captain  Kater,  of  the  Briti^fc 
Imperial  Troy  pound.'  Report  of  the  Secretary  of  the  Treasury,  March  3,  1831, 

7* 


148  COMPOUND    NUMBERS.  [SECT.  VIII 

equal  to  32  grains,  was  called  a  pennyweight,  from  its  being  the  weight  of  the 
silver  penny  then  in  circulation.  At  a  later  period  the  pennyweight  was  di- 
vided into  24  equal  parts  instead  of  3*2,  which  are  still  called  grains,  being  the 
smallest  weight  now  in  common  use. 

Note. — Troy  Weight  was  formerly  used  in  weighing  articles  of  every  kind, 
It  was  introduced  into  Europe  from  Cairo  in  Egypt,  about  the  time  of  the 
Crusades,  in  the  12th  century.  Some  suppose  its  name  was  derived  from 
Troyes,  a  city  in  France,  which  first  adopted  it ;  others  think  it  was  derived 
from  Troy-novant)  the  former  name  of  London.* 


AVOIRDUPOIS   WEIGHT. 

251*  Avoirdupois  Weight  is  used  in  weighing  groceries  ana 
all  coarse  articles ;  as,  sugar,  tea,  coffee,  butter,  cheese,  flour,  hay, 
&c.,  and  all  metals  except  gold  and  silver. 

16  drams  {dr.)  make  1  ounce,             marked  oz. 

16  ounces  "     1  pound,                  "  Ib. 

25  pounds  "     1  quarter,                "  qr. 

4  quarters,  or  100  Ibs.  "     1  hundred  weight,  "  cwt. 

20  hundred  weight  "     1  ton,                       "  T. 

Note. — In  weighing  wool  in  England,  7  pounds  make  1  clove ;  2  cloves,  1 
stone;  2 stone,  1  tod;  6fc  tods,  Iwey;  2weys,  1  sack;  12  sacks,  1  last;  240 
pounds,  1  pack. 

OBS.  1.  Formerly  it  was  the  custom  to  allow  112  pounds  for  a  hundred 
weight,  and  28  pounds  for  a  quarter ;  but  this  practice  has  become  nearly  or 
juite  obsolete.  In  buying  and  selling  all  articles  of  commerce  estimated  by 
weight,  the  laws  of  most  of  the  States  as  well  as  general  usage,  call  100 
pounds  a  hundred  weight,  and  25  pounds  a  quarter. 

2.  Gross  weight  is  the  weight  of  goods  with  the  boxes,  casks,  or  bags  which 
contain  them. 

Net  weight  is  the  weight  of  the  goods  only. 

252*  The  Avoirdupois  Pound  of  the  United  States,  is  equal 
to  27.701554  cubic  inches  of  distilled  water,  at  the  maximum 
density,  and  at  30  inches  barometer.f  It  is  determined  from  the 
Troy  Pound,  by  the  legal  proportions  of  5760  grains,  which  con- 

QUEST.— 251.  In  what  is  Avoirdupois  Weight  used  ?  Repeat  the  Table  1  CV;s.  How  majiy 
pounds  were  formerly  allowed  for  a  hundred  weight  ?  For  a  quarter  ?  What  is  gross  weight! 
Net  weight  7  252.  How  is  the  Avomhipois  pound  of  the  United  States  determined  ? 

*  Hind's  Arithmetic,  Art.  224.    Also,  North  American  Roview,  Vol  XLV. 
t  Reports  of  Secretary  of  Treasury,  March  3,  1832:  June  30,  1832.    Also,  Congressional 
iKtcaments  ot  1833. 


ARTS.  251-253.  J       COMPOLND  NUMBERS.  149 

jstitute  the  Troy  pound,  to  7000  grains  Troy,  which  constitute  the 
Avoirdupois  pound.     That  is, 

5700  grains  Troy  make  1  pound  Troy. 

7000  grains               "  "      1  pound  Avoirdupois. 

437^  grains               "  "      1  ounce  " 

27i£  grains               "  "      1  dram  " 

OBS.  1,  The  British  Imperial  Pound  Avoirdupois  is  equal  to  27-7274  cubic 
inches  of  distilled  water,  at  the  temperature  of  62°  Fahrenheit,  when  the 
barometer  stands  at  30°.  It  is  determined  from  the  Imperial  Troy  pound, 
which  contains  57(>0  grains,  while  the  former  contains  7000  grains. 

2.  Since  the  Troy  pound  of  the  United  States  is  identical  with  the  Troy 
pound  of  England,  the  Avoirdupois  pound  of  the  former  must  be  equal  to  that 
of  the  latter  ;  for  both  bear  the  same  ratio  to  the  Troy  pound.     But  the  Eng- 
lish avoirdupois  pound  is  said  to  contain  27.7274  cu.  in.  of  distilled  water, 
while  that  of  the  United  States,  according  to  Mr.  Hassler,  contains  27.701554 
cu.  in.     This  slight  difference  may  be  accounted  for  by  the  fact  that  the  for- 
mer was  measured  at  the  temperature  of  02°,  while  the  latter  was  measured 
at  its  maximum  density,  which  is  39.83  degrees. 

3.  The  standard  of  weight  adopted  by  the  State  of  New  York,  in  1827,  is  the 
avoirdupois  pound,  whose  magnitude  is  such  that  a  cubic  foot  of  distilled 
water,  at  the  maximum  density,  in  a  vacuum,  will  weigh  62£  pounds,  or  1000 
ounces. 

Note. — The  term  avoirdupois,  is  thought  by  some  to  be  derived  from  the 
French  avoir  du  poids,  a  phrase  signifying  to  have  weight.  Others  think  it 
is  from  avoirs,  the  ancient  name  of  goods  or  c/tattels,  and  poids  signifying 
weight  in  the  Norman  dialect.* 

APOTHECARIES'   WEIGHT. 

253.  Apothecaries'  Weight  is  used  by  apothecaries  and  phy- 
sicians in  mixing  medicines. 

20  grains  (^r.)  make  1  scruple,  marked  sc.,   or  3. 

3  scruples                    "      1  dram,  "      dr.,  or  3. 

8  drams                        "      1  ounce,  "       oz.,  or  5. 

12  ounces                      "      1  pound,  "  tt>. 

OBS.  1.  The  pound  and  ounce  in  this  weight  are  the  same,  as  the  Troy 
pound  an<\  ounce;  the  other  denominations  are  different. 

2.  Drugs  and  medicines  are  bought  and  sold  by  avoirdupois  weight. 

QricsT.— 2.13.  In  what  is  Apothecaries'  Weight  used?  Recitt  the  Table?  Ohs.  To 
What  are  the  aix>thecaries'  ounce  and  pound  equal  ?  How  are  drugs  and  medicines  bought 
and  soldi 

*  President  John  Qaincy  Adams  on  Weights  and  Measures ;  also,  Hind's  Arithmetic. 


150  COMPOUND    NUMBER8.  [SECT.   VIII 


LONG  MEASURE. 

254*  Long  Measure  is  used  in  measuring  distances  where 
length  only  is  considered,  without  regard  to  breadth  or  depth. 
It  is  frequently  called  linear  or  lineal  measure. 

12  inches  (in.)  make  1  foot,  marked  ft. 

3  feet  "     1  yard,  "  yd. 

5j  yards,  or  16£  feet  "     1  rod,  perch,  or  pole,     "  r.  or  p. 

40  rods  "     1  furlong,  "  fur. 

8  furlongs,  or  320  rods        "     1  mile,  "  m. 

3  miles  "     1  league,  "  I. 

60  geographical  miles,  or  |    „     j  d  „  ^          „. 

G9£  statute  miles  S 

360  degrees  make  a  great  circle,  or  the  circumference  of  the  earth. 

Note. — 4  inches  make  1  hand ;  9  inches,  1  span ;  18  inches,  1  cubit ;  6  feet 
1  fathom. 

In  measuring  roads  and  land,  surveyors  use  a  chain  which  is  4  rods  long,  and 
which  is  divided  into  100  links.  Hence,  25  links  make  1  rod,  and  7-f2^  inches 
make  1  link.  This  chain  is  commonly  called  G-unter's  Chain,  from  the  name 
of  its  inventor. 

OBS.  1.  The  inch  is  commonly  divided  either  into  eighths  or  tenths;  some- 
times, however,  it  is  divided  into  twelfths,  which  are  called  lines.  Formerly 
the  inch  was  divided  into  3  barleycorns ;  but  the  barleycorn  is  not  employed  as 
a  measure  at  the  present  day.  The  term  barleycorn,  is  derived  from  a  grain  of 
barley,  which  was  the  original  element  of  Linear  Measure. 

2.  The  terms  rod,  pole,  and  -n^ch,  from  the  French  perche  signifying  a  rod 
are  each  expressive  of  the  instrument,  which  was  originally  used  as  a  measure 
of  this  length. 

255.  The  standard  Unit  of  Length  adopted  by  the  Unite 
States,  is  the  Yard  of  3  feet,  or  36  inches,  and  is  identical  with 
the  British  Imperial  Yard.  It  is  made  of  brass,  at  the  temper- 
ature of  62°  Fahrenheit,  from  the  scale  of  iighty-two  inches  pre- 
\.  pared  by  Trough  ton,  a  celebrated  English  artist,  for  the  survey 
of  the  Coast  of  the  United  States. 

OBS.  1.  The  Imperial  standard  yard  of  Great  Britain  is  determined  fromtha 
end-idum  which  vibrates  seconds  in  a  vacuum,  at  the  level  of  the  sea.  IB 

QTTEST.— 254.  In  what  is  Long  Measure  used?  What  is  Long  Measure  sometimes 
called?  Recite  the  Table?  Obs.  How  are  inches  usually  divided?  What  is  the 
origin  of  the  measure  called  barleycorn  ?  Is  this  measure  now  used  ?  255.  \V  hat  is  tb» 
standard  unit  of  Length  adopted  by  the  United  States  1 


ARTS.  254-257.]       COMPOUND  NUMBERS.  151 

Greenwich  or  London.  This  pendulum  is  divided  into  391393  equal  parts 
aid  300000  of  these  parts  are  declared,  by  act  of  Parliament,  to  be  the  stand- 
ard yard,  at  the  temperature  of  62° ;  consequently,  since  the  yard  is  divided 
into  36  inches,  it  follows  that  the  length  of  a  pendulum  vibrating  seconds,  un- 
der these  circumstances,  is  39.1393  inches. 

The  English,  yard  is  stud  to  have  been  originally  determined  by  the  length 
of  the  arm  of  Henry  I.  King  of  England. 

2.  The  standard  of  linear  measure  adopted  by  the  State  of  New  York,  is 
the  pendulum  which  vibrates  seconds,  in  a  vacuum,  at  Columbia  College,  in  the 
city  of  New  York,  which  is  in  the  latitude  of  40°  42',  43".  The  yard  is  de- 
clared to  be  -KHjiHHH"  of  this  pendulum  ;  hence,  the  length  of  the  .pendulum 
is  39.101688  inches,  at  the  temperature  of  32°.  Should  the  standard  yard 
ever  be  lost,  it  could  be  recovered  by  resorting  to  the  preceding  experiment 

CLOTH   MEASURE. 

2 56*  Cloth  Measure  is  used  in  measuring  cloth,  lace,  and  all 
kinds  of  goods,  which  are  bought  and  sold  by  the  yard. 

2$  inches  (in.}  make  1  nail,  marked  na. 

4  nails,  or  9  in.  "     1  quarter  of  a  yard,       "       qr. 

4  quarters  "     1  yard,  "       yd. 

3  quarters,  or  f  of  a  yard    "     1  Flemish  ell,  "       PI.  e. 

5  quarters,  or  1  j  yard         "     1  English  ell,  "       E.  e. 

6  quarters,  or  l£  yard         "     1  French  ell,  "       F.  e. 

OBS.  Cloth  measure  is  a  species  of  long  measure.  Cloth,  laces,  &c.,  are 
bought  and  sold  by  the  linear  yard,  without  regard  to  their  width. 

SQUARE  MEASURE. 

257*  Square  Measure  is  used  in  measuring  surfaces,  or 
things  whose  length  and  breadth  are  considered  without  regard  to 
height  or  depth ;  as,  land,  flooring,  plastering,  &c. 

144  square  inches  (sq.  in.}  make  1  square  foot,  marked  sq.ft. 

9  square  feet  "     1  square  yard,  "  sq.  yd. 
30|  square  yards,  or  ) 

272i  square  feet           \  l  **'  rod>  Perch> or  Pole>  s?' r' 

40  square  rods  1  rood,  "  R. 

4  roods,  or  160  square  rods   "     1  acre,  "  A. 

640  acres  "     1  square  mile,  "  M. 

QTTEST.— 256.  In  what  is  cloth  measure  used  ?  Repeat  the  Table  Obs.  Of  what  \M 
cloth  measure  a,  species  1  What  is  the  kind  of  yard  by  which  cloths,  laces,  &c ,  are  bough} 
and  sold  ?  257.  In  what  is  Square  Measure  used  ?  Recite  the  Table. 


152 


COMPOUND    NUMBERS. 


[SECT.  VIIL 


Note. — 16  square  rods  make  1  square  chain  ;  10  square  chains,  or  100,000 
square  links,  make  an  acre.  Flooring,  roofing,  plastering,  &c.,  are  frequently 
estimated  by  the  "  square,"  which  contains  JOO  square  feet. 

A  hide  of  land,  which  is  spoken  of  by  ancient  writers,  is  100  acres. 

OBS.  1.  A.  square  is  a  figure  which  h&sfmir  equal  sides,  and  all  its  angles 
right  angks,  as  seen  in  the  diagram.  Hence, 

A  Sqibare  Inch  is  a  square,  whose  sides  are  each  a      9  sq.ft.=\  sq.  yd. 

linear  inch  in  length. 

A  Square  Foot  is  a  square,  whose  sides  are  each  a 
<#aeai  foot  in  length. 

A  Square  Yard  is  a  square,  whose  sides  are  each  a 
linear  yard,  or  three  linear  feet  in  length,  and  con- 
tains 9  square  feet,  as  represented  in  the  adjacent 
figure. 

2.  Square  measure  is  so  called,  because  its  measuring  unit  is  a  square.  Tho 
standard  of  square  measure  is  derived  from  the  standard  linear  measure.  Hence, 

A  unit  of  square  measure  is  a  square  whose  sides  are  respectively  equal,  in 
length,  to  the  linear  unit  of  the  same  name. 


CUBIC  MEASURE. 


258.  Cubic  Measure  is  used  in  measuring  solid  bodies,  01 
things  which  have  length,  breadth,  and  thickness  ;  such  as  timber, 
stone,  boxes  of  goods,  the  capacity  of  rooms,  ships,  &c. 


1728  cubic  inches  (cu.  tw.)  make  1  cubic  foot, 


27  cubic  feet 
40  feet  of  round,  or 
50  ft.  of  hewn  timber 
42  cubic  feet 

16  cubic  feet 

8  cord  feet  or 
128  cubic  feet 


1  cubic  yard, 
1  ton,  or  load, 

1  ton  of  shipping, 
1  foot  of  wood,  or  ) 
a  cord  foot,         ) 

1  cord, 


marked  cu.ft. 
"        cu.  yd. 


c. 


A  pile  of  wood  8  feet  long,  4  feet  wide,  and  4  feet  high,  contains  1 
For,  8X4X4;=  128. 


»ru, 


.—  Obs.  What  is  a  square  1    What  \r  a  square  inch  ?     A  square  foot  1    A  square 
258.  In  what  is  Cubic  Measure  used       Recite  the  Table. 


ARTS.  258,  259.]       COMPOUND  NUMBERS. 


153 


/ 


_-  I1-?; 
£v-;   -    -  "~"~r-^ 


Obs.  1.  A  Cube  is  a  solid  body  bounded  by  six 
equal  sides.  It  is  often  called  a  hexaedron.  Hence? 

A  Cubic  Inch  is  a  cube,  each  of  whose  sides 
is  a  square  inch,  as  represented  by  the  adjoin- 
ing figure. 

A  Cubic  Foot  is  a  cube,  each  of  whose  sides 
is  a  square  foot. 

2.  Cubic  Measure  is  so  called,  because  its  meas- 
uring unit  is  a  cube.  It  is  often  called  solid  meas- 
ure.    The  standard  of  cubic  measure  is  derived  from  the   standard  linear 
measure. 

A  unit  of  cubic  measure,  therefore,  is  a  cube  whose  sides  are  respectively 
equal  in  length  to  the  linear  unit  of  the  same  name. 

3.  The  cubic  ton,  sometimes  called  a  load,  is  chiefly  used  for  estimating  the 
cartage  and  transportation  of  timber.    By  a  ton  of  round  timber  is  meant, 
such  a  quantity  of  timber  in  its  rough  or  natural  state,  as  when  hewn,  will 
make  40  cubic  feet,  and  is  supposed  to  be  equal  in  weight  to  50  feet  of  hewn 
timber. 

The  cubic  ton  or  load,  is  by  no  means  an  accurate  or  uniform  stand- 
ard of  estimating  weight ;  for,  different  kinds  of  timber,  are  of  very  different 
degrees  of  density.  But  it  is  perhaps  sufficiently  accurate  for  the  purposes  to 
which  it  is  applied. 

Note. — For  an  easy  method  of  forming  models  of  the  Cube  and  other  regular 
Solids,  see  Thomson's  Legendre's  Geometry,  p.  222. 

WINE   MEASURE. 

259*  Wine  Measure  is  used  in  measuring  wine,  alcohol,  mo- 
lasses, oil,  and  all  other  liquids  except  beer,  ale,  and  milk. 


4  gills  (gi.) 

2  pints 

4  quarts 
31  &  gallons 
42  gallons 
63  gallons,  or  2  barrels 

2  hogsheads 

2  pipes 


marked  pt. 


make  1  pint, 
"     1  quart, 
"        gallon, 
"        barrel, 
"     1  tierce 

hogshead, 

pipe  or  butt, 

tun, 

OBS.  1.  In  England,  10  gallons  make  1  anker;  18  gallons,  1  runlet;  9 
tierces  or  84  gallons,  1  puncheon. 

2.  Liquids  are  generally  bought  and  sold  by  the  gallon  or  its  subdimnons,  a* 

QUKST.--O&S.  What  is  a  cube  ?  What  is  a  cubic  inch  ?  A  cubic  foot  ?  What  is 
meant  by  a  ton  of  round  timber  ?  259.  In  what  is  Wine  Measure  used  ?  Recite  the 
Table. 


bar.  or  bbl. 

tier. 

Mid. 

pi. 

tun. 


154  COMPOUND    NUMBERS.  [StiCT.  VI11  , 

the  quart,  pint,  &c.  Cider  and  a  few  cheap  articles  are  bought  and  sold  by 
the  barrel.  The  capacities  of  cisterns,  vats,  &c  are  sometimes  estimated  hi 
hogsheads,  and  the  quotations  or  prices-current  of  oils  in  foreign  markets,  are 
usually  made  in  tuns.  But  the  tierce,  and  the  pipe  or  bull  are  never  used,  as 
such,  in  business  transactions;  their  contents  are  given  in  gallons,  quarts,  &c. 

26O.  The  standard  Unit  of  Liquid  Measure  adopted  by  the 

1     United  States,  is  the  Wine  Gallon  of  231  cubic  inches,  which  is 

equal  to  58372.175^  grains  of  distilled  water,  at  the  maximum 

density,  weighed  in  air  at  30   inches  barometer,   or    8.339  Ibs. 

avoirdupois,  very  nearly.* 

OBS.  The  British  imperial  standard  measure  of  capacity,  both  for  fluids  and 
dry  goods,  is  the  imperial  gallon,  which  is  equal  to  10  pounds  avoi  -upois  of 
distilled  water,  at  02°  thermometer  and  30  inches  barometer,  ai»  contains 
277.274  cubic  inches.  It  is  equal  to  1.2  gal.  wine  measure  U.  S. 

BEER  MEASURE. 
26 1  •  Beer  Measure  is  used  in  measuring  beer,  ale,  and  milk 

2  pints  (pis.}  make  1  quart',  marked  qt. 

4  quarts  "  1  gallon,  "  gal. 

36  gallons  "  I  barrel,  "  bar.  or  bbl 

1  i  barrels,  or  54  gallons  "  1  hogshead,        "  lihd. 

OBS.  1.  In  England,  9  gallons  make  1  frrkin;  2  firkins,  1  kilderkin;  2  kil 
derkins,  1  barrel. 

2.  The  beer  gallon  contains  282  cubic  inches,  and  is  equal  to  10.1799321 
pounds  avoirdupois  of  distilled  water,  at  the  maximum  density.  In  many  place* 
milk  is  measured  by  wine  measure. 

DRY  MEASURE. 

262*  Dry  Measure  is  used  in  measuring  grain,  fruit,  cfec. 

2  pints  (pf.)  make  1  quart,  marked  qt. 

8  quarts  "  1  peck,  "  pk. 

4  pecks,  or  32  qts,  «  1  bushel,  "  bu. 

8  bushels  "  1  quarter,  "  qr. 

32  bushels,  or  4  qrs.  "  1  chaldron,  "  cli. 

QUEST.— 260.  What  is  the  standard  unit  of  Liquid  Measure  of  the  Baited  Sta/es? 
How  many  cubic  inches  in  a  wine  gallon  ?  261.  In  what  is  Beer  Measure  ise«l  1  Recite 
the  Table.  262.  In  what  is  Dry  Measure  used  1  Repeat  the  Table. 

*  Reports  of  the  Secretary  of  the  Treasury,  March,  1831,  and  June,  '  832.  A  so,  Hassle* 
on  Weights  and  Measures. 


ARTS.  260-263.]      COMPOUND  NUMBERS.  155 

OBS.  In  England  flour  is  often  sold  by  weight.  A  sack  is  equal  to  280 
lb«?.,  and  contains  about  five  imperial  bushels. 

The  following  denominations,  are  sometimes  used,  viz :  2  quarts  make  1 
pottle;  2  bushels,  1  strike;  2  strikes  or  4  bu.,  1  coom;  2  cooms  or  8  bu.,  1 
quarter  ;  5  quarters,  1  wey  or  load  ;  2  loads,  1  last. 

In  London  36  bushels  of  coal  make  a  chaldron,  but  in  New  Castle  79J 
bushels  are  said  to  be  allowed  for  a  chaldron.  But  coal  in  England  aud  in 
this  country,  is  now  usually  bought  and  sold  by  weight. 

Note. —  Wine,  Beer,  and  Dry  Measures  are  often  called  capacity  measure* 
and  are  evidently  a  species  of  cubic  measure. 

263.  The  standard  Unit  of  Dry  Measure  adopted  by  th 
United  States,  is  the  Winchester  Hushel,  which  is  equal  t 
77.027413  pounds  avoirdupois  of  distilled  water,  at  the  max 
imum  density,  weighed  in  air  at  30  inches  barometer,  and  contain* 
2150.4  cubic  inches,  nearly. 

The  Winchester  bushel  is  so  called,  because  the  standard  meas 
ure  was  formerly  kept  at  WincJiester,  England.     By  statute,  it  '4 
an  upright  cylinder,  18^  inches  in  diameter,  and  8  inches  deep. 

OBS.  1 .  The  imperial  bushel  of  Great  Britain  is  equal  to  80  Ibs.  avoirdupois 
of  distilled  water,  at  62°  Fahrenheit,  and  30  inches  barometer,  and  contains 
2218. 192  cubic  inches;  consequently,  it  is  equal  to  1.032  bushel  U.  S.,  nearly. 
It  is  an  upright  cylinder,  whose  internal  diameter  is  18.789  inches,  and  its 
depth  8  inches. 

The  use  of  heaped  measure  was  abolished  by  Act  of  Parliament,  in  1835. 

2.  The  standard  bushel  of  the  State  of  New  York,  is  equal  to  80  pounds 
avoirdupois  of  distilled  water,  at  the  maximum  density,  at  the  mean  pressure 
of  the  atmosphere,  and  contains  2218.192  cubic  inches.* 

It  is  customary,  at  the  present  day,  to  determine  capacity  measures  by  the 
weight  of  distilled  water  which  they  contain.  This  is  evidently  nv  re  accu- 
rate than  the  former  method  of  measurement  by  cubic  inches. 

3.  In  buying  and  selling  grain,  when  no  special  agreement  as  to  measure- 
ment or  weight,  is  made  by  the  parties,  a  bushel,  in  the  State  of  New  York,  by 
Act  of  1836,  consists  of  60  Ibs.  of  wheat,  56  Ibs.  rye  or  Indian  corn,  48  Ibs.  cf 
barley,  and  32  Ibs.  of  oats. 

There  are  similar  statutes  in  most  of  the  other  States  of  the  Union.  This  is 
the  most  impartial  method  by  which  the  value  of  grain  can  be  estimated. 

QUEST. — 063.  What  is  the  standard  unit  of  Dry  Measure  adopted  by  the  Government  1 

*  By  the  same  Act  it  was  declared,  that  the  standard  liquid  gallon  should  be  8  Ibs.,  and 
the  standard  dry  gallon  10  Ibs.  avoirdupois  of  distilled  water,  at  its  maximum  density 
Butttis  part  of  the  statute  was  subsequently  repealed,  and  the  previous  standard  gallon 
In  the  office  of  the  Secretary  of  State,  was  continued  m  use. 


156  COMPOUND    NUMBERS.  [SECT.    VIII 

TIME. 

204.  Time  is  naturally  divided  into  days  and  years ;  the  for- 
mer are  caused  by  the  revolution  of  the  Earth  on  its  axis,  the 
latter  by  its  revolution  round  the  sun. 

60  seconds  'see.)  make  1  minute,            marked  min. 

60  minutes  "  1  hour,  "  hr. 

24  noun  "  1  day,  "  d. 

7  days  "  1  week  «  wk. 

4  weeks  "  1  lunar  month,       "  mo. 

12  calendar  months,  or  ) 

365  days  and  6  hrs,  (nearly,)  \  1  civil  year>  *• 

The  following  are  the  names  of  the  12  calendar  months  into  which  the  civj 
or  legal  year  is  divided,  with  the  number  of  days  in  each. 

January,  written  (Jan.)  the  first        month,  has  31  days. 

February,  "  (Feb.)  "  second  "  "  28  " 

March,  "  (Mar.)  "  third  "  "  31  « 

April,  "  (Apr.)  "  fourth  "  "  30  « 

May,  «  (May)  «  fifth  «  «  31  « 

June,  "  (June)  "  sixth  "  "  30  « 

July,  "  (July)  "  seventh  «  "  31  " 

August,  "  (Aug.)  «  eighth  «  "  31  " 

September,  "  (Sept.)  "  ninth  "  "  30  " 

October,  "  (Oct.)  "  tenth  "  "  31  " 

November,  ""  (Nov.)  "  eleventh  "  "  30  " 

December,  "  (Dec.)  "  twelfth  "  "  31  « 

The  number  of  days  m  each  month  may  be  easily  remembered  from  the  fol- 
lowing lines: 

"  Thirty  days  hath  September, 
April,  June,  and  November ; 
February  twenty-eight  alone, 
All  the  rest  have  thirty-one  ; 
Except  in  Leap  year,  then  is  the  time, 
When  February  has  twenty-nine." 

OBS.  1.  A  Solar  year  is  the  exact  time  in  which  the  earth  revolves  round 
the  sun,  and  contains  365'days,  5  hours,  48  minutes,  and  48  seconds. 

2.  Since  the  civil  year  contains  365  days  and  6  hours,  (nearly,)  it  is  plain 
iha,  in  four  years  a  whole  da/  will  be  gained,  and  therefore  every  fourth  year 
must  have  366  days.  This  day  was  originally  added  to  the  year,  by  repeating  the 
sixth  of  the  Calends  of  March  i\  the  Roman  calendar,  which ,  corresponds  with 

QUEST.— 264.  'low  is  time  natural  7  divided  ?  Recite  the  Table.  Obs  What  is  a  solai 
year  ?  How  is  leap  year  occasioned*.  To  which  month  is  the  odd  day  added] 


ARTS.  264-266.]       COMPOUND  NUMBERS,  157 

the  24th  of  February  in  ours.  It  was  called  the  interculary  day,  from  the 
Latin  intercalo,  to  insert. 

The  year  in  whch  this  day  is  added,  is  called  Bissextile,  from  the  Latin  bis, 
twice,  and  sextilis,  the  sixth.  It  is  also  called  "  Leap  Year"  because  it  leapa 
over  a  day  more  than  a  common  year. 

3.  The  civil  or  kgal  year  is  often  called  the  Julian  year,  from  Julius  Caesar 
emperor  of  Rome,  who  adapted  the  calendar  or  register  of  the  civil  year  to 
the  supposed  lengtl  of  the  solar  year,  oy  adding  1  day  to  every  fourth  year. 

265*  In  process  of  time,  as  mathematical  and  astronomica 
science  advanced,  it  was  found  that  the  length  of  a  solar  yeai 
was  only  365  d.  5  hrs.  48min.  48  sec.,  or  11  min.  12  sec.  less 
than  365-}-  days,  which  in  400  years  amounted  to  about  3  days ; 
consequently,  the  Julian  calendar  was  behind  the  solar  time. 
This  error  at  the  time  of  Pope  Gregory  XIII.,  amounted  to  10 
days,  which  he  corrected  in  1582  by  suppressing  10  days  in  the 
month  of  October,  the  day  after  the  4th  being  called  the  15th. 
Hence  this  calendar  is  sometimes  called  the  Gregorian  calendar. 

OBS.  1.  This  correction  was  not  adopted  in  England  till  1752,  when  the 
'error  amounted  to  11  days.  By  Act  of  Parliament,  11  days,  after  the  2d  of 
September,  were  therefore  omitted ;  and  the  civil  year  by  the  same  Act,  was 
made  to  commence  on  the  1st  of  January,  instead  of  the  25th  of  March,  as  it 
had  done  previously. 

2.  Dates  reckoned  by  the  old  method  or  Julian  calendar,  are  called  Old 
Style ;  and  those  reckoned  by  the  new  method,  are  called  New  Style. 

To  change  any  date  from  Old  to  New  Style,  we  must  add  1 1  days  to  it ;  and 
if  the  given  date  in  Old  Style,  is  between  the  1st  of  January  and  the  25th  of 
March,  we  must  add  1  to  the  year  in  New  Style. 

Russia  still  reckons  dates  according  to  Old  Style.  The  difference  now 
amounts  to  12  days. 

!£(>(•>.  To  ascertain  whether  a  year  is  LEAP  YEAK. 

Divide  the  given  year  by  4,  ana  if  there  is  no  remainder,  it  is 
Leap  year.  The  remainder,  if  any,  shows  how  many  years  have 
elapsed  since  a  Leap  year  occurred.  Thus,  dividing  the  year  1847 
by  4,  the  remainder  is  3  ;  hence  it  is  3  years  since  the  last  leap 
year,  and  the  ensuing  year  will  be  leap  year. 

OBS.  1.  To  this  rule  there  is  an  exception.  For,  we  have  seen,  that  a  solzi 
year  is  11  min.  and  12  sec.  less  than  a  Julian  year,  which  is  365$  days,  Tnia 
«rror,  in  400  years,  amounts  to  about  3  days ;  consequently,  if  1  day  is  added 

QUEST.— 266.  How  do  you  ascert  <in  whether  a  year  is  leap  year? 


158 


COMPOUND    NUMBERS. 


[SECT.  VIII 


every  fourth  year ,  that  ia,  if  we  have  100  leap  years  in  400  years,  according  to 
the  Julian  calendar,  the  reckoning  would  fall  3  days  behind  the  solar  time. 
Thus  reckoning  from  the  commencement  of  the  Christian  era,  when  it  was 
January  1st,  401  by  the  Julian  time,  it  was  January  4th  by  the  solar  time. 

2.  To  remedy  this  error  only  1  centennial  year  in  four  is  regarded  a  kap 
year ;  or,  which  is  the  same  in  effect,  whenever  the  centennial  year,  or  the 
number  expressing  the  century,  is  not  divisible  by  4,  that  year  is  not  a  leap 
year,  while  the  other  centennial  years  are.  Thus,  17,  18,  19,  denoting  1700, 
1800,  and  1900,  are  not  divisible  by  4,  consequently  they  are  not,  leap  yaars, 
though  according  to  the  rule  above  they  would  be  ;  on  the  other  hand  1 6  and 
20,  denoting  1600  and  2000,  are  divisible  by  4,  and  are  therefore  leap  years. 
There  is  still  a  slight  error,  but  it  is  so  small  that  in  5000  years  it  scarcely 
amounts  to  a  day. 

CIRCULAR  MEASURE,   OR  MOTION. 

267.  Circular  Measure  is  applied  to  the  divisions  of  the  cir- 
cle, and  is  used  in  reckoning  latitude  and  longitude,  and  the 
motion  of  the  heavenly  bodies. 

60  seconds  (")  make  1  minute,  marked 

60  minutes  "  1  degree,  "  o 

30  degrees  "  1  sign,  «  5. 

12  signs,  or  360°  "  1  circle,  "  c. 

This  measure  is  often  called  Angular  Measure,  and  is  chiefly  usec  by 
astronomers,  navigators,  and  surveyors. 


OBS.  1.  The  circumference  of  every  cir- 
cle is  divided  or  supposed  to  be  divided, 
into  360  equal  parts,  called  degrees,  as  in 
the  subjoined  figure. 

2.  Since  a  degree  is  -^^  part  of  the 
circumference  of  a  circle,  it  is  obvious  that 
its  length  must  depend  on  the  size  of  the 
circle. 


270° 


Note.— The  division  of  the  circumference  of  the  circle  into  36C  equal  part*, 
took  its  origin  from  the  length  of  the  year,  which,  (in  round  numbers)  wa* 
supposed  to  contain  360  days,  or  12  months  of  30  days  each.  The  12  signs 

QUXST  —267.  In  what  is  Circular  Measure  used  ?  Repeat  the  Table.  Obs.  How  is  the 
circumference  of  every  circle  divided  ?  On  what  does  the  length  of  a  degree  depend  1 


ARTS.  267-269.]       COMPOUND  NUMBERS. 


159 


corresponl  to  the  12  months.     Thfc  \errn  minutes,  is  from  the  Latin 

which  signifies  a  small  part.     The  term  seconds,  is  an  abbreviated  expression 

for  second  minutes,  or  minutes  of  the  second  order. 

268.  Since  the  earth  turns  on  its  axis  from  west  to  east  once 
in  24  hours,  it  evidently  revolves  15°  per  hour;  or  1°  in   4  min- 
utes, and  1'  in  4  seconds  of  time.     Hence, 

When  the  difference  of  longitude  between  two  places  is  V,  the 
difference  in  the  time,  or  the  hour  of  the  day  at  these  two  places,  is  4 
teconds  ;  if  the  difference  of  longitude  is  1° ,  the  difference  of  time  u 
4  minutes  ;  if  2°,  the  difference  of  time  is  8  minutes,  &c. 

Thus,  when  it  is  noon  at  London,  in  Philadelphia,  which  is  about 
75°  west  from  London,  it  is  only  7  o'clock,  A.  M.  For,  if  the 
earth  revolves  1°  in  4  minutes,  to  revolve  75°,  it  will  require  75 
times  as  long,  and  4x75  =  300  min.,  or  5  hours. 

OBS.  1.  Since  the  earth  revolves  from  west  to  east,  it  is  manifest,  that  the 
time  is  earlier  as  we  go  eastward,  and  later  as  we  go  westward. 

2.  This  principle  affords  navigators  and  others  a  convenient  and  useful 
method  of  ascertaining  the  difference  of  time  between  two  places,  when  the 
difference  of  their  longitude  is  known  ;  also,  for  ascertaining  the  difference  of 
longitude  between  two  places,  when  the  difference  in  their  time  is  known. 

MISCELLANEOUS    TABLE. 

269.  The  following  denominations  no-   included  in  the  pre- 
ceding Tables,  are  frequently  used. 

12  units    ' 

12  dozen,  or  144 

12  gross,  or  1728 

20  units 

50  pounds 
100  pounds 

30  gallons 

200  Ibs.  of  shad  or  salmon 
19(!  pounds 
200  pounds 

14  pounds  of  iron  or  lead 
21  i  stone 


make 


dozen,  (doz?) 


great  gross. 

score. 

firkin  of  butter. 

quintal  of  fish. 

bar.  of  fish  in  Mass. 

bar.  in  N.  Y.  and  Conn. 

bar.  of  flour. 

bar.  of  pork. 

stone. 


1  fother. 


Note.  —  Formerly  it  was  customary  to  allow  112  Ibs.  for  a  quintal. 


QPKST.— 268.  When  the  difference  of  longitude  between  two  r laces  is  1',  what  Is  UM 
iifference  of  time  7    Wher  1°,  what  is  the  difference  of  time  ? 


160 


COMPOUND    NUMBERS 


[SECT.  VIII. 


PAPER  AND   BOOKS. 

27CK  The  terms  folio,  quarto,  octavo,  &3.,  applied  to  books, 
denote  the  number  of  leaves  into  which  a  sheet  of  paper  is 
bided. 


24  sheets  of  paper 
20  quires 

2  reams 

5  bundles 

A  sheet  folded  in  two  leaves 


make 


A  sheet 
A  sheet 
A  sheet 
A  sheet 
A  sheet 


"  four  leaves 
"  eight  leaves 
"  twelve  leaves 
"  eighteen  leaves 
"  thirty-six  leaves 


1  quire. 
1  ream. 
1  bundle. 
1  bale. 


forms    a 


a  quarto,  or  4to. 
an  octavo,  or  8vo. 
a  duodecimo,  or  12mo. 
an  18mo. 
a  36mo. 


DIMENSIONS    OF    DIFFERENT    KINDS    OF    ENGLISH    PAPER. 


Names. 

Writing. 

Drawing. 

Printing. 

Pott, 

15*  by  12*  in. 

15*  by  12*  in. 

Small  Post, 

16*  by  13*  in. 

Fool's  Cap, 

16|  by  13*  in. 

16f  by  13*  in. 

Crown, 

20  by  15  in. 

20  by  15  in. 

Demy, 

20  by  15*  in. 

22  by  17  in. 

Medium, 

22*  by  17*  in. 

23  by  18  in. 

Royal, 

24  by  19*  in. 

24  by  19i  in. 

26  by  20  in. 

Super  Royal, 

27*  by  19i  in. 

27*  by  19i  in. 

• 

Elephant, 

28  by  23  in. 

Double  Crown, 

30  by  20  in. 

Imperial, 

30}  by  22  in. 

30j  by  22  in. 

Atlas, 

34  by  26£  in. 

Columbier, 

34*  by  23*  in. 

Double  Demy, 

38*  by  26  in. 

Double  Elephant, 

40  by  26}  in. 

Antiquarian, 

52  by  31  in. 

Double  Atlas, 

55  by  31*  in. 

Emperor, 

* 

68  by  48  in. 

Note. — American  paper  is  usually  rather  larger  than  English  paper  at  tie 
game  name. 


QUEST. — 270.  What  do  the  terms,  folio,  quarto,  &c.,  denote,  when  applied  to  books  I 
What  is  a  folio  1    A  quarto  ?    An  octavo  1    A  duodecimo  ?    An  18r»o.  7    A  36mo.  5 


ARTS.  270-273.]      COMPOUND  NUMBERS.  1   1 

FRENCH  MONEY,   WEIGHTS,   AND   MEASURES. 

271.  The  new  system  of  Money,  Weights,  and  Measures  of 
France,  adopted  in   1795,  was  formed  according  to  the  decimal 
Notation. 

FRENCH    MONET. 

272.  The  Franc  is  the  unit  money  of  the  new  system   of 
French  currency.     It  is  a  silver  coin,  consisting  of  -fa   pure  sil- 
ver, and  iV  of  alloy. 

10  centimes  make  1  declme. 

4  10  declines  "  1  franc. 

2V0fe. — The  value  of  a  franc  by  Act  of  Congress  in  1843,  is  $.186.  The 
value  of  the  hvre  lournois,  the  former  unit  of  money,  is  $.185. 

FRENCH    LINEAR    MEASURE. 

273.  The  standard  unit  of  the  French  Linear  Measure,  is 
the  Metre.    Its  length,  according  to  the  mean  of  the  several  com- 
parisons of  Troughton,  Nicollet  and  Hassler,  is  equal  to  39. 3809171 
English,  or  United  States  inches. 

10  metres        make  1  decametre  =  32.817431  U.  S.  feet. 

10  decametres      "  1  hectometre  =  328.17431      "       « 

10  hectometres    "  1  kilometre  =  3281.7431      "       " 

10  kilometres       "  1  myriametre  =  32817.431      "       " 

Note. — 1.  The  standard  by  which  the  new  French  measures  of  length  are 
determined,  is  the  quadrant  of  a  meridian  of  the  earth,  or  the  terrestrial  arc 
from  the  equator  to  the  pole,  in  the  meridian  of  Paris.  The  ten-millionth\  part 
of  this  arc  is  called  a  metre,  which  is  equal  to  39.378  U.  S.  in.,  nearly. 

2.  The  metre  is  divided  into  10  decimetres ;  the  decimetre  into  10  centimetres ; 
the  centimetre  into  10  millimetres. 

3.  The  denominations  of  the  old  system  of  linear  measure  were  the  toise, 
foot,  inch,  line,  and  point.     12  points=l  line;  12  lines=:l  inch;  12  in.=l 
foot;  6  ft.=l  toise.     The  old  French  foot  was  equal  to  1.066  U.  S.  feet. 

*.  By  a  decree  of  1812,  the  Toise,  Aune,  Foot,  &c.,  are  allowed  to  be  used, 
having  the  following  ratios  to  the  metre,  viz:  the  toise =2  metres ;  the  foot= 
J-  metre ;  the  inch=-l-L2-  metre ;  the  aune  or  ell=:l£  metre ;  the  bushel—  £  hec- 
tolitre. 


162  COMPOUND    CUMBERS.  [SECT.    VI ll. 


FRENCH    SQUARE    MEASURE. 

274.  The  unit  of  French  Superficial  Measure,  is  the  Are, 
Miose  sides  are  each  a  decametre  in  length ;  consequently,  it  con- 
*-iins  100  square  metres,  or  119.6648496  U.  S.  sq.  yds. 

10  ares       make  1  decare  =  1196.64849G  U.  S.  sq.  yds. 

tOdecares      "  1  hectare  =11966.48496     "        " 

10  hectares    "  1  kilare  =  119664.8496     "        " 

lOkilares       "  1  myriare  =1196648.496     "        " 

Note. — The  are  is  divided  into  10  declares;  the  declare  into  10  centiares; 
the  centiare  into  10  milliares. 

» 
FRENCH  CUBIC    MEASURE. 

275.  The  unit  of  French  Cubic  Measure,  is  the  Stere,  which 
is  a  cubic  metre,  and  is  equal  to  61074.1564445  cu.  in.  U.  S. 

10  decisteres  make  1  stere         =  35.34384  cu.  ft.  U.  S. 
10  steres  "      1  decastere  =  353.4384      "        " 

FRENCH    LIQUID    AND    DRY    MEASURE. 

276.  The  unit  of    French   Liquid  and   Dry  Measures,  is 
called   the   Litre,  which   is   a   cubic  decimetre,  and  is    equal  to 
61.0741564445  cu.  in.  U.  S.,  or  1.05756  qts.  wine  measure. 

10  litres        make  1  decalitre  =  2.6439  gals,  wine  meas. 
10  decalitres      "    1  hectolitre  =  26.439     "        "      " 
10  hectolitres    "    1  kilolitre    =  264.39    "        "      " 

Note. — The  litre  is  divided  into  10  decilitres ;  the  decilitre  into  10  centilitres; 
the  centilitre  into  10  millilitres. 

FRENCH    WEIGHTS. 

277.  The  unit  of  French  Weights,  is  the  weight  of  a  cubic 
centimetre  of  distilled  water,   at  the  maximum  density,    and  is 
called  the  Gramme.     It  is  equal  to  15.433159  grains  Troy. 

10  grammes        make  1  decagramme    =  154.33159  grs.  Troy. 
10  decagrammes     "      1  hectogramme  =  1543.3159      "       " 
10  hectogrammes    "      1  kilogramme     =  15433.159      "       " 
10  kilogrammes,      "      1  myriagramme  =  154331.59      "       " 


Avrs.  275-279.]     FOREIGN   WEIGHTS,  ETC.  163 


.  —  I.  The  gramme  is  divided  into  10  decigrammes,  the  dctigramrfie 
into  10  centigrammes  ;  the  centigramme  into  10  milligrammes. 

""2.  The  denomination  chiefly  used  in  making  out  invoices  of  goods  sold  by 
weight,  and  in  business  transactions,  is  the  Idhgramme,  which  is  equal  to  1000 
grammes,  or  2.21  Ibs.  avordupois,  very  nearly. 

3.  In  the  old  system  of  French  weight,  the  livre-poids—  2  marcs;  the  marc 
=8  onces  ;  the  once—  8  gros  ;  the  gros=72  grains.  The  livre  is  equal  to  one- 
half  the  kilogramme. 

FRENCH    CIRCULAR    MEASURE. 

278.  The  c/rc/e  is  divided  into  400  equal  parts,  called  grades, 
and  the  quadrant  into  100  grades.     The  grade  is  again  divided 
into  100  equal  parts,  and  each  of  these  parts  is  subdivided  into 
100  other  equal  parts,  according  to  the  centesimal  scale.     Hence, 

The  seconde     =     .00009  English  deg. 
The  minute      =     .009  "  « 

The  grade        =     .9  "  " 

Note.  —  The  names  of  the  denominations  larger  than  the  unit  in  the  French 
Compound  Numbers,  are  formed  by  prefixing  to  the  name  of  the  unit,  the 
Greek  words,  dcca,  heclo,  kilo,  and  myria;  those  less  than  the  unit,  are  formed 
by  prefixing  to  the  name  of  the  unit,  the  Latin  words,  aeci,  centi,  and  milLL 

279.  Foreign  Weights  and  Measures  compared  with  those  of 
the  United  States* 

Amsterdam.—  100  Ibs.  (1  centner)=108.923  Ibs.f;  1  last=85.25  bu.;  1  ahm= 
41  gals.;  1  foot  Amsterdam—  11-f  in.;  1  foot  Antwerp  =11-}-  in.  ;  1  ell  Am- 
sterdam=2.26  ft.;  1  ell  Brabant=2.3ft.  ;  1  ell  Hague=2.28  ft. 

Daiai'ia.  —  1  picul=l36  Ibs.;   1  kann=.39  gal.;   1  ell=2.25  ft. 

Bengal.  —  I  haut=l.5  ft.;  1  guz=3  ft.;  1  coss  or  mile  =1.24  miles;  1  bazar 
maud=82.14  Ibs.  ;  1  factory  maud=74.G(5  Ibs. 

Bencoolen.  —  1  bahar=560  Ibs.;   1  bamboo  =1  gal.;  1  coyangr=8  gals. 

Bombay.  —  1  maud=28  Ibs.  ;   1  covid=1.5  ft.  ;  1  candy=25  bu. 

Bremen.—  1  pound  =1.1  lb.;  1  centner^  116  Ibs.  ;  1  last=80.7  bu.  ;  1  ft.=ll|in 

Canton.  —  1  tael  =  l}  oz.  ;  1  catty  =U  Ibs.  ;  1  picul=l33$  Ibs.  ;  1  covid—  14|  in. 

Denmark.—  100  Ibs.  (1  centner)  =  110.25  Ibs.;  1  bbl.  (toende)=3.95  bu.; 
1  \iertel=2.04  gals.  ;  1  foot  Copenhagen,  or  Rhineland=12}  in. 

Florence  and  Leghorn.  —  100  Ibs.  (1  cantaro)=74.8(j  Ibs.  ;  1  moggio=  16.59  bu.; 
1  barile=12.04  gals.;  1  palmo=9f  in. 

*  M'Culloch's  Commercial  Dictionary;  also  Kelly's  Universal  Cambist. 
t  The  poutvls  in  this  and  the  following  comparisons  are  avoirdupois. 
T.H.  8 


164  FOREIGN    WEIGHTS,    ETC.  [SECT.  VI11. 

Genoa. — 100  \')s.  (1  peso  grosso)=76f  Ibs. ;  t  peso  sottile=69.89  Ibs. ;  1  oina 

=3.43  bu. ;  1  mezzarola=39.22  gals. ;  1  palmo=9-£-  in. 
Hamburg*—!  foot=l  1.3  hi. ;  1  ell =22.6  in.  nearly;  1  ell  Brabant=27.6  in. ; 

1  mile=4.68  miles ;  lfass=Hbu. ;  1  last=89.64  bu. ;   1  ahm=38i  gals. 
Japan.— I  catti=1.3  Ibs.;  1  picul=l30  Ibs. ;  1  ichan=3£  ft.;  1  inc  or  tetamy 

=6^  ft.;  1  balec=16i  gals. 
Madras.— 1  covid=l£  ft.;    1  catty =1J  Ibs.;  1  picul=l33|  Ibs.;  1  maud= 

25  Ibs.;  1  candy =500  Ibs.;  1  garee=140bu. 

Malta.—  1  foot=10-£-  in.;  100  Ibs.  (1  cantaro)= 174.5  Ibs. ;  1  salma— 8.22  bu, 
Manilla. — 1  arroba=26  Ibs. ;  1  picul=143  Ibs. ;  1  palmo=  10.38  in. 
Naples. — 1  cantaro  grosso=196.5  Ibs.;  1  cantaro  piccolo=l06  Ibs.;  1  palmo=a 

10$  in ;  1  tomolo=1.45  bu. ;  1  carro=52.24bu. ;  1  carro  of  wine =264  gals. 
Netherlands.— 1  ell =3.28  ft.;    1  mudde=2.84  bu.;    1  kan  litre =2. 11  pints; 

1  vat  hectolitre=26.42  gals. ;  1  pond  kilogramme— 2.21  Ibs. 
Portugal.— 100  lbs.  =  101.19  Ibs.;  1  arroba=22.:26  Ibs. ;  1  quintal =89.05  Ibs. ; 

1  almude=4.37gals.;  1  alquiere=4f  bu. ;  1  moyo=23.03  bu.;  1  last=70  bu. ; 

1  peor  foot=l2-k  in.;  1  mile=l?  mile. 
Prussia.— 100  lbs=103.11  Ibs. ;    1  quintal  (110  lbs.)=H3.42  Ibs. ;  1  eimar= 

18.14  gal.;  1  scheffel=1.56  bu. ;  1  foot=1.03  ft. ;  I  ell=2.l9ft.;  1  mile=: 

4.68  miles. 
Rome.—lQQ  libras=74.77  Ibs. ;  1  rubbio=9.3G  bu. ;  1  barile=15.3l  gals. ;  1  foot 

=llfin.;  1  canna=(>£  ft- ;   I  mile=7|  fur. 
Russia— 100  lbs.=90.26  Ibs.;    1  berquit=361.04  Ibs.;    40  Ibs.  (1  pood)= 

36  Ibs. ;  1  vedro=3i  gals. ;  1  chetwert=5.95  bu. ;  1  foot  Petersburg  1.18  ft.t; 

1  foot  Moscow=l.l  ft;  1  arsheen=2£  ft.;  1  mile  (verst)=5.3  fur. 
Sicily. — 100  Ibs.  (libras)=70  Ibs.;    1  cantaro  grosso=192.5  Ibs.;   1  cantarf 

sottile=l75  Ibs.;  1  salma  generale=7.85  bu. ;    1  salma  grossa=9.77  bu. ; 

1  salma  of  wine=23.06  gals. ;  1  palmo=9£  in. ;  1  canna=6£  ft. 
Spain. — 1  arroba=25.36  Ibs.;    1  quintal=lOl.44  Ibs.;    1    arroba  of  wine = 

4|  gals. ;  1  moyo=G8  gals. ;  1  fanega  =1.6bu. ;  1  foot=ll.l28  in.;  1  vara= 

2.78  ft. ;  1  league  (leagua)=4.3  m.,  nearly. 
Sweden.— 100  Ibs,  (  -ictualie) =73.76   Ibs. ;  "l  foot=ll.G9  in. ;    1  ell=1.95  ft.; 

1  mile=6.64  m. ;    1  kann=7.42  bu. ;    1  last=75  bu. ;    1  kann  of  wine= 

69.09  gals. 
Smyrna— 100  Ibs.  (1  quintal) =129.48  Ibs.;    1   oke=2.83  Ibs.;        quillot^ 

1.46  bu. ;  1  quillot  of  wine=l3.5  gals. ;  1  pic=2j-  ft. 
Trieste.— 100  Ibs.=123.6  Ibs  ;  1  stajo=2i  bu.;  1  orna,  or  eimer=1494  gals.; 

1  ell  (for  silk)=:2.1  ft. ;  1  ell  (for  woollen)=2.2  ft. ;  1  foot  Austrian=  1.037  ft.  ; 

1  mile  Austrian=4.6  m. 
1  enice.— 100  Ibs.  (Ipesso  grosso)=105.l8  Ibs. :  1  peso  sottile=64.42  Ibs, ;  1  staja 

=2.27  bu.;  1  moggio=9.08bu.;  1  anifora= 137  gals.;  1  foot=1.14ft.;  1  brae- 

cio  (for  silk)  =24.8  in. ;  1  braccio  (for  woollen) =26.6  in. 

*  New  system  of  weights  and  measures  adopted  in  1843. 

t  In  measuring  timber  English  feet  and  inches  are  chiefly  used  throughout  Russia. 


ART.  280.]  REDUCTION.  165 


REDUCTION. 

28O»  The  process  of  changing  compound  numbers  from  one 
denomination  into  another,  without  altering  their  value,  is  called 
REDUCTION. 

Ex.  1.  Reduce  £5,  2s.  Yd.  and  3  far.  to  farthings. 

Analysis. — Since  in  £l  there  are  20s.,  in  £5  there  are  5  times 
as  many,  which  is  100s.,  and  2,  (the  given  shillings,)  make  102s. 
Again,  since  there  are  12d.  in  Is.,  in  102s.  there  are  102  times 
as  many,  which  is  equal  to  1224d.,  and  7  (the  given  pence)  make 
123 Id.  Finally,  since  in  Id.  there  are  4  far.,  in  123 Id.  there  are 
1231  times  as  many,  or  4924  far.,  and  3,  (the  given  far.,)  make 
4927  far.  Ans.  4927  farthings. 

Operation. 

£     s.     d.  far.  We  first  reduce  the  given  pounds  to  shil- 

5273.  lings,  by  multiplying  them  by  20,  because 

20s.  in  £1.  20s.  make   £l.    (Art.  247.)     We  next  re- 

102  shillings.  duce  the  shillings  to  pence,  by  multiply- 

12d.  in  Is.  ing  them  by  12,  because  12d.  make  Is.     Fi- 

1231  pence.  nally>  we  reduce  t C  pome  to  farthings  by 

4  far  in  Id.  multiplying    them    by    4,    because    4    far. 

4927  far.     Ans.  make  Id. 

Note. — 1.  In  this  example  it  is  required  to  reduce  higher  denominations  to 
lower ;  as  pounds  to  shillings,  shillings  to  pence,  &c.  This  is  done  by  suc- 
cessive multiplications. 

2.  In  4927  farthings,  how  many  pounds,  shillings,  and  pence? 

Analysis. — Since  4  far.  make  Id.,  in  4927  farthings,  there  are 
as  many  pence  as  4  is  contained  times  in  4927,  which  is  123 Id., 
and  £  far.  over.  Again,  since  12d.  make  Is.,  in  1231d.  there  are 
as  many  shillings  as  12  is  contained  times  in  1231,  which  is 
102s.,  and  7d.  over.  Finally,  since  20s.  make  £l,  in  102s.  there 

QUEST. — 280.  What  is  Reduction  7  How  are  pounds  reduced  to  shillings  ?  Why  mul 
tiply  by  20  ?  How  are  shillings  reduced  to  pence  ?  Why  1  How  pence  to  farthings  ?  Whv 


166  REDUCTION.  [SlJCT.    VIIL 

are  as  many  pounds  as  20  is  contained  times  in  102,  which  .'is  £5, 
and  2s.  over.  Am.  £5,  2s.  Yd.  3  far. 

Operation.  We  first  reduce  the  given  farthings 

4)4927  far.  to  pence,  the  next  higher  denomina- 

12)  123 Id.  3  far.  over.       tion,  by  dividing  them  by  4,  becaus« 

20)  102s.  7d.  over.  4  far.  make  Id.  (Art.  247.)    Next  we 

£5,  2s.  over.  reduce  the  pence  to  shillings  by  di- 

Ans.    £5,  2s.  7d.  3.  far.    viding    them    by    12,    because    12d. 

flLake  1  s.     Finally,  we  reduce  the  shillings  to  pounds  by  dividing 

them  by  20,  because  20s.  make  £l.     The  last  quotient  with  the 

several  remainders,  constitute  the  answer. 

Note. — 2.  The  last  example  is  exactly  the  reverse  of  the  first ;  that  is,  lower 
denominations  are  reduced  to  higher,  which  is  done  by  successive  divisions. 

281*  From  the  preceding  illustrations  we  derive  the  fol- 
lowing 

GENERAL  RULE  FOR  REDUCTION. 

I.  To  reduce  compound  numbers  to  lower  denominations. 
Multiply  the  highest  denomination  given,  by  that  number  which 

it  takes  of  the  next  lower  denomination  to  make  ONE  of  this  higher  ; 
to  the  product,  add  the  number  expressed  in  this  lower  denomina- 
tion in  the  given  example.  Proceed  in  this  manner  with  each 
successive  denomination,  till  you  come  to  the  one  required. 

II.  To  reduce  compound  numbers  to  higher  denominations. 
Divide  the  given  denomination  by  that  n  imber  which  it  takes  of 

this  denomination  to  make  ONE  of  tlie  next  higher.  Proceed  in  this 
manner  with  each  successive  denomination,  till  you  come  to  the  one 
tequired.  The  last  quotient,  with  tJie  (everal  remainders,  will  be 
the  answer  sought. 

282»  PROOF. — Reverse  the  operation  /  that  is,  reduce  back  the 
answer  to  the  original  denominations,  and  if  the  result  corresponds 
with  the  numbers  given,  the  work  is  right. 

QUEST. — How  are  farthings  reduced  to  pence  ?  Why  divide  by  4  ?  How  reduce  pence 
to  shillings?  Why?  How  reduce  shillings  to  pounds  ?  Why?  281.  How  are  compound 
numbers  reduced  to  lower  denominations?  How  to  higher  denominations?  282  How 
is  Reduction  proved  ? 


ARTS.  281,  282. 1  REDUCTION.  167 

OBS.  1.  Each  remainder  is  of  the  same  denomination  as  the  dividend  from 
which  it  arose.  (Art.  113.  Obs.  1.) 

2.  Reducing  compound  numbers  to  lower  denominations  may,  with  propri- 
ety, be  called  Reduction  by  Multiplication ;  reducing  them  to  higher  denom- 
inations, Reduction  by  Division.  The  former  is  often  called  Reduction  De* 
vending ;  the  latter,  Reduction  Ascending.  They  mutually  prove  each  other. 

EXAMPLES    FOR   PRACTICE. 

I.  In  136  rods  and  2  yards,  how  many  feet? 

Operation.  Proof, 

rods.  yds.  3)2250  ft. 

2)136  2  5i)750  yds. 

__H  yds.  1  r.  _^ 

682  11)1500 

68  .        T36f  r.  4  rem.=2  yards. 

750  yds.  Now  136  r.  2  yds.  is  the 

3  ft.  1  yd.  given  number. 
2250  ft.    Ans. 

2.  In  £71,  13s.  6£d.,  how  many  farthings? 

3.  In  £90,  Vs.  8d.,  how  many  farthings  ? 

4.  In  £295,  18s.  3|d.,  how  many  farthings? 

5.  In  95  guineas,  17s.  9fd.,  how  many  farthings? 

6.  How  many  pounds,  shillings,  &c.,  in  24651  farthings? 

7.  How  many  pounds,  shillings,  &c.,  in  415739  farthings? 

8.  How  many  guineas,  &c.,  in  67256  pence? 

9.  In  £36,  4s.,  how  many  six-pences  ? 

10.  In  £75,  12s.  6d,,  how  many  three-pences  ? 

II.  Reduce  29  Ibs.  7  oz.  3  pwts.  to  grains. 

12.  Reduce  37  Ibs.  6  oz.  to  pennyweights. 

13.  Reduce  175  Ibs.  4  oz.  5  pwts.  7  grs.  to  grains. 

14.  Reduce  12256  grs.  to  pounds,  &c. 

15.  Reduce  42672  pwts.  to  pounds,  <fec. 

16.  In  15  cwt.  3  qrs.  21  Ibs.,  how  many  pounds? 

17.  In  17  tons  12  cwt.  2  qrs.,  how  many  ounces? 

QUEST. — Obs.  Of  what  denomination  is  each  remainder  7  What  may  reducing  compound 
numbers  to  lower  denominations  be  called?  To  higher  denominational  Which  of  ih« 
fundamental  rules  la  employed  by  the  formef  ?  Which  by  the  latter  1 


168  REDUCTION.  [SECT. 

18.  In  52  tons  3  cwt.,  how  many  pounds  ? 

19.  In  140  tons,  how  many  drams  ? 

20.  In  16256  ounces,  how  many  hundred  weight,  <fec.  ? 

21.  In  267235  pounds,  how  many  tons,  <fec.? 

22.  In  5637 2 8  drams,  how  many  tons,  pounds,  <fec.  ? 

23.  Reduce  95  pounds  (apothecaries'  weight)  to  drams. 

24.  Reduce  130  pounds  to  scruples. 

25.  Reduce  6237  drams  (apothecaries'  weight)  to  pounds,  <feu 

26.  Reduce  25463  scruples  to  ounces,  &c. 

27.  How  many  feet  in  27  miles  ? 

28.  How  many  inches  in  45  leagues  ? 

29.  How  many  yards  in  3000  miles  ? 
30    In  290375  feet,  how  many  miles? 

31.  In  1875343  inches,  how  many  leagues? 

32.  In  15  m.  5  fur.  31  r.,  how  many  rods  ? 

33.  In  1081080  inches,  how  many  miles,  &c.  ? 

34.  How  many  feet  in  the  circumference  of  the  earth  ? 

35.  How  many  nails  in  160  yards? 

36.  How  many  quarters  in  1000  English  ells? 

37.  In  102345  nails,  how  many  yards,  &c.  ? 

38.  In  223267  nails,  how  many  French  ells? 

39.  In  634  yards,  3  qrs.,  how  many  nails  ? 

40.  In  28  hhds.  15  gals,  wine  measure,  how  many  quarts? 

41.  In  5  pipes,  1  hhd.,  how  many  gallons? 

42.  In  3  tuns,  1  hhd.  10  gals.,  how  many  gills? 

43.  In  12256  pints,  how  many  barrels,  wine  measure? 

44.  In  475262  gills,  how  many  pipes,  &c.  ? 

45.  In  50  hhds.  1  bbl.  10  gals.,  how  many  gills,  wine  measure . 

46.  In  45  bbls.,  how  many  pints,  beer  measure  ? 

47.  How  many  barrels  of  beer  in  25264  pints  ? 

48.  How  many  hogsheads  of  beer  in  136256  quarts? 

49.  How  many  pints  in  45  hhds.  10  gals,  of  beer  ? 
60.  In  15  bushels,  1  peck,  how  many  quarts? 

51.  In  763  bushels,  3  pecks,  how  many  quarts? 

52.  In  56  quarts,  5  bushels,  how  many  pints  ? 

53.  In  45672  quarts,  how  many  bushels,  &e.  ? 
64.  In  260200  pints,  how  many  quarts  ? 


ARTS.  282,  283. J  REDUCTION.  169 

55.  Reduce  25  days,  6  hours  to  minutes. 

56.  Reduce  365  days,  6  hours  to  seconds. 

57.  Reduce  847125  minutes  to  weeks. 
58  Reduce  5623480  seconds  to  days. 
59.  How  many  seconds  in  a  solar  year? 

GO.  How  many  seconds  in  30  years,  allowing  365  days  6  houri 
to  a  year? 

61.  How  many  years  of  Sabbaths  are  there  in  70  years? 

62.  In  110  degrees,  20  minutes,  how  many  seconds?     . 

63.  In  11  signs,  45  degrees,  how  many  seconds? 

64.  In  7654314  seconds,  how  many  degrees? 

65.  In  1000000000  minutes,  how  many  signs? 

66.  Reduce  1728  sq.  rods,  23  yds.  5  feet  to  feet. 

67.  Reduce  100  acres,  37  rods  to  square  feet. 

68.  Reduce  832590  sq.  rods  to  sq.  inches. 

69.  Reduce  25363896  sq.  feet  to  acres,  &c. 

70.  In  150  cubic  feet,  how  many  inches? 

71.  In  97  yds.  15  ft.,  how  many  cubic  inches? 

72.  In  49  cords,  23  feet,  how  many  cubic  inches? 

73.  In  84673  cubic  inches,  how  many  feet  ? 

74.  In  39216  cubic  feet,  how  many  cords? 

75.  In  65  tons  of  round  timber,  how  many  cubic  inches? 

76.  In  4562100  cubic  inches,  how  many  tons  of  hewn  timber? 

APPLICATIONS   OP   REDUCTION. 

283.  To  reduce  Troy  to  Avoirdupois  weight. 

First  reduce  the  yiven  pounds,  ounces,  d'c.,  to  grains  ;  then  divide 
oy  the  number  of  grains  in  a  dram,  and  t/te  quotient  will  be  t/ie  an- 
swer in  drams.  (Art.  252.) 

OBS.  If  the  answer  is  required  to  be  in  pounds  and  a  fraction  of  a  j  iu^4 
divide  the  grains  !)y  7000. 

Ex.  1.  In  175  pounds  Troy,  how  many  pounds  avoirdupois? 
Solution.— 175X12X20X24  =  1008000    grs.,    and    1008000 
grs.-~27ii=36864  drams,  or  144  Ibs.  avoirdupois.     Ans. 

"  — — - —  > 

How  is  Troy  weight  reduced  to  avoirdupois  1 


170  REDUCTION.  [SECT.  VIIL 

2.  In  700  Ibs.  Troy  of  silver,  how  many  pounds  avoirdupois  ? 

3.  In  840  Ibs.  6  oz.  10  pwts.,  how  many  pounds,  &c.,  avoirdu- 
pois ? 

4.  An  apothecary  bought  1000  Ibs.  of  opium  by  Troy  weight, 
and  sold  it  by  avoirdupois  :  how  many  pounds  did  he  lose  ? 

6.  A  merchant  bought  1500  pounds  of  lead  Troy  weight,  and 
sold  it  by  avoirdupois  :  how  many  pounds  did  he  lose  ? 

284*  To  reduce  Avoirdupois  to  Troy  weight. 

First  reduce  the  given  pounds,  ounces,  &c.y  to  drams,  then  multiply 
ty  the  number  of  grains  in  a  dram,  and  the  product  will  be  the  an- 
swer  in  grains.  (Art.  252.) 

OBS.  1.  When  the  given  example  contains  pounds  only,  we  may  multiply 
them  by  7000,  and  the  product  will  be  grains. 

2.  If  the  answer  is  required  to  be  in  pounds  and  a  fraction  of  a  pound,  di- 
vide the  grains  by  57GO. 

6.  In  32  Ibs.  avoirdupois,  how  many  pounds  Troy  ? 
Solution.  —  32X16Xl6X27ii=224000  g1"8-*  and  224000  grs. 

=38  Ibs.  10  oz.  13  pwts.  8  grs.     Ans. 

7.  In  48  Ibs.  avoirdupois,  how  many  pounds  Troy  ? 

8.  A  merchant  bought  100  Ibs.  10  oz.  of  tea  avoir.,  and  sold  it 
by  Troy  weight  :  how  many  pounds  did  he  gain  ? 

9.  A  druggist  bought  1260  Ibs.  of  alum  avoirdupois,  and  re- 
tailed it  by  Troy  weight  :  how  many  more  pounds  did  he  sell 
than  he  bought  ? 

285.  The  area  of  a  floor,  a  piece  of  land,  or  any  surface  which 
has  four  sides  and  four  right-angles,  is  found  by  multiplying  its 
length  and  breadth  together 

Note  1.  The  area  of  a  figure  is  the  superficial  contents  or  space  contained 
within  the  line  or  lines,  by  which  the  figure  is  bounded.  It  is  reckoned  in 
square  inches,  feet,  yards,  rods,  &c. 

2.  A  figure  which  has  four  sides  and  four  right-angles,  like  the  following 
diagram,  is  called  a  Rectangle  or  Parallelogram. 


.—  284.  How  is  avoirdupois  weight  reduced  to  Troy?  285.  How  do  you  find 
the  area  or  superficial  contents  of  a  surface  having  four  side*  and  four  right-angle*  1 
JVbte.  What  is  meant  by  the  term  area  7  How  is  it  reckoned  1  What  is  a  figure  wnich 
lias  four  sides  and  four  right-angles  called  ? 


ARTS.  284-286.]  REDUCTION.  171 

10.  How  many  square  yards  of  carpeting  will  it  take  to  cover 
a  room,  4  yards  long  and  3  yards  wide  ? 

Suggestion. — Let  the  given  room  be 
represented  by  the  subjoined  figure,  the 
length  of  which  is  divided  into  4  equal 
pans,  and  the  breadth  into  3  equal 
parts,  which  we  will  call  linear  yards. 
'Now  it  is  plain  that  the  room  will  con- 
tain as  many  square  yards  as  there  are 
squares  in  the  given  figure.  But  the 
numbe  of  squares  in  the  figure  is  equal  to  the  number  of  equal 
parts  (linear  yards)  which  its  length  contains,  repeated  as  many 
times  as  there  are  equal  parts  (linear  yards)  in  its  breadth ;  that 
is,  it  is  equal  to  4X3,  or  12.  Ans.  12  yds. 

11«  How  many  sq.  feet  in  a  floor,  20  feet  long,  18  feet  wide  ? 

12.  How  many  acres  in  a  field,  50  rods  long,  45  rods  wide  ? 

13.  How  many  square  yards  in  a  ceiling,  35  feet  long  and  28 
feet  wide  ? 

14.  How  many  acres  in  a  farm,  420  rods  long  and  170  rods 
wide  ? 

15.  What  is  the  area  of  a  square  field,  whose  sides  are  80  rods 
in  length  ? 

16.  How  many  yards  of  carpeting,  a  yard  wide,  will  it  take  to 
cover  a  floor  18  feet  square. 

17.  How  many  yards  of  plastering  are  required  to  cover  four 
sides  of  a  room,  18  ft.  long,  15  feet  wide,  and  9  ft.  highj? 

18.  How  many  square  yards  of  shingling  will  cover  both  sides 
of  a  roof,  whose  rafters  are  20  feet,  and  whose  ridge  pole  is  25 
feet  long  ? 

286.  The  epical  contents,  or  solidity  of  boxes  of  goods, 
piles  of  wood,  &c.,  are  found  by  multiplying  the  length,  breadth, 
and  thickness  together. 

19.  How  many  cubic  feet  in  a  box  5  feet  long,  ^  feet  wide  and 
3  feet  deep  ? 

Solution. — 5X4=20,  and  20X3=60.     Ans.  60  cu.  ft. 


Q0S8T.— 286.  How  are  the  cubical  contents  of  a  box  of  goods,  a  pile  of  wood  %e. 
found  1 

8* 


172  REDUCTION  [SECT.  VIII 

20.  How  many  cubic  feet  in  a  Mock  of  granite,  65  in.  long,  42 
in.  wide,  and  36  in.  thick? 

21.  How  many  cubic  feet  in  a  load  of  wood,  8  ft.  long,  4^  ft. 
high,  and  3£  ft.  wide  ? 

22.  How  many  cords  of  wood  in  a  pile,  46  ft.  long,  16  ft.  high, 
and  14if|  feet  wide? 

23.  How  many  cubic  feet  in  a  vat,  12  ft.  long,  8£  ft.  wide,  and 
?i  ft.  deep  ? 

24.  How  many  cubic  feet  in  a  bin,  12  ft.  long,  9  ft.  deep,  and  7 
ft.  wide  ? 

25.  How  many  cubic  yards  in  a  cellar,  18  ft.  long,  12  ft.  wide, 
and  9  ft.  deep  ? 

26.  How  many  cubic  feet  in  a  stick  of  timber,  2  ft.  square,  and 
40  ft.  long  ? 

27.  How  many  cubic  feet  in  a  cistern  15  ft.  long,  12  ft.  wide, 
and  10  ft.  deep  ? 

287.  To  reduce  Cubic  to  Dry,  or  Liquid  Measure. 

First  reduce  the  given  yards,  feet,  <kc.,  to  cubic  inches ;  then 
divide  by  tJie  number  of  cubic  indies  in  a  gallon,  or  bushel,  as 
the  case  may  be,  and  the  quotient  will  be  the  answer  required 
(Arts.  260,  263.) 

28.  In  10752  cubic  feet,  how  many  bushels? 

Solution.  —  10752  X  1728  =  18579456  cubic  inches;  and 
185794564-2150-^=8640  bushels. 

29.  In  21504  cubic  feet,  how  many  bushels  ? 

30.  In  462  cubic  feet,  how  many  wine  gallons  ? 

31.  In  1155  cubic  feet  and  33  inches,  how  many  ^.vine  gallons  ? 

32.  In  846  cubic  feet,  how  many  beer  gallons  ? 

33.  In  1128  cubic  feet  and  141  in.,  how  many  beer  gallons  ? 

34.  How  many  bushels  will  a  bin  contain,  which  is  5  ft.  long, 
6  ft.  wide,  and  4  ft.  deep  ? 

35.  How  many  bushels  will  a  bin  contain,  which  is  8  ft.  long, 
4f  ft.  wide,  and  3i  ft.  deep  ? 

QUEST — 287.  How  reduce  cubic  to  dry,  or  liquid  measure  ? 


ARTS.  287-289.  J  REDUCTION.  173 

36.  How  many  bushels  will  a  bin  contain,  which  is  14  ft.  long, 
10  ft.  8  in.  wide,  and  6  ft.  8  in.  deep  ? 

37.  How  many  wine  gallons  in  a  cistern,  which  is  6  ft.  long, 
5  ft.  wide,  and  4  feet  deep  ? 

38.  How  many  barrels  of   water  (wine  meas.)  will  a  cistern 
hold,  which  is  20  ft.  long,  15  ft.  wide,  and  10  ft.  deep  ? 

39.  The  distributing  reservoir  of  the  Croton  Water  Works  in 
the  City  of  New  York,  is  436  ft.  square  and  40  feet  high:  how 
many  hogsheads  of  water  will  it  hold  ? 

288.       To  reduce  Dry,  or  Liquid,  to  Cubic  Measure. 

First  find  the  number  of  bushels,  if  dry  measure,  or  gallons,  if 
liquid  measure,  in  the  given  example  j  then  multiply  by  the  num- 
ber of  cubic  inches  in  a  gallon,  or  bushel,  as  the  case  may  be,  and 
the  product  will  be  the  answer  required.  (Art.  263.) 

40.  How  many  cubic  feet  in  a  bin,  which  contains  100  bushels  ? 

Solution.  —  100x2150-14ir  =  215040,     and    215040  -J-  1728= 
,  or  124£  cubic  feet.     Ans. 


41.  How  many  cubic  feet  in  a  lime  kiln,   which  holds  100 
bushels  ? 

42.  How  many  cubic  feet  in  the  hold  of  a  ship,  which  contains 
1000  bushels  of  grain  ? 

43.  How  many  cubic  feet  in  1  hogshead,  wine  measure? 

44.  How  many  cubic  feet  in  a  cistern,  which  holds  50  barrels 
of  water? 

45.  How  many  cubic  feet  in  a  vat,  which  contains  100  hogs- 
heads wine  measure  ? 

289.  To  reduce  Liquid  to  Dry  Measure,  or  Dry  to  Liquid 
Measure. 

First  find  the  cubic  inches  in  tlw  given  example  ;  then  divid 
them  by  the  number  of  cubic  inches  in  a  gallon,  or  bus\elt  as  t/te  case 
may  be,  and  the  quotient  will  be  the  answer  required. 

QUIBT.—  288.  II  >w  reduce  dry,  or  liquid  measure  to  cubic?     289   How  reduce  liquid 
to  dry  measure  ?    How  dry  to  liquid  measure  1 


174  REDUCTION.  [SECT.  YII1 

46.  In  40  gallons  wine  measure,  how  many  bushels  ? 
Solution.— 40X  231=9240  cu.  in.,  and  9240  cu.  in. H-2150-r*ff= 

4-JH  bushels.     Ans. 

47.  In  6  hogsheads,  16  gallons,  how  many  bushels  ? 

48.  In  5  bushels,  how  many  gallons  wine  measure  ? 

49.  In   3200  quarts  dry  measure,  how  many  hogsheads  wine 
measure  ? 

29O.  To  reduce  Wine  ti  Beer  Measure,  or  Beer  to  Wine 
Measure. 

First  find  the  number  of  cubic  inches  in  the  the  given  example  ; 
then  divide  them  by  tlie  number  of  cubic  inches  which  it  takes  to 
make  a  gallon  in  tJie  required  measure. 

50.  In  94  wine  gallons,  how  many  beer  gallons  ? 

Solution. — 94X231  =  21714  cu.  in.,  and  21714  cu.  in. —282  = 
77  gallons.  Ans. 

51.  In  1  hhd.  wine  measure,  how  many  beer  gallons? 

52.  A  tavern-keeper  bought  4  hhds.  of  cider  wine  measure,  and 
retailed  it  by  beer  measure  :  how  many  gallons  did  he  lose  ? 

53.  In  20  beer  gallons,  how  many  wine  gallons? 

54.  A  grocer  bought  7238  gallons  of  milk  beer  measure,  and 
retailed  it  by  wine  measure :  how  many  gallons  did  he  gain  ? 

55.  A  druggist  bought  10000  gallons  of  alcohol  beer  measure, 
and  sold  it  by  wine  measure :  how  many  gallons  did  he  gain  ? 

56.  A  grocer  bought  65  hhds.  29  gals,  and  2  quarts  of  milk  by 
Deer  measure,  and  sold  it  to  his  customers  by  wine  measure :  how 
many  quarts  more  did  he  sell  than  he  bought  ? 

57.  A  liquor  dealer  bought  120  pipes  of  wine  which  his  clerk 
retailed  by  beer  measure :  how  many  gallons  more  did  he  buy 
than  he  sold  ? 

291  •  Since  the  earth  revolves  on  its  axis  1°  in  four  minutes, 
or  I'  in  4  seconds  of  time,  (Art.  268,)  it  is  evident  that  longitude 
nay  be  reduced  to  time.  That  is,  multiplying  degrees  of  longi- 
tude by  4  reduces  them  to  minutes  of  time,  multiplying  minutes 
of  longitude  by  4  reduces  them  to  seconds  of  time,  &c. 

QDKST. — 290.  How  reduce  wine  to  beer  measure  7    How  oeer  to  wine  measure  7 


ARTS.  290-293.  J  REDUCTION.  115 

By  reversing  this  process  it  is  evident  that  time  may  be  reduced 
co  longitude.  Thus,  dividing  seconds  of  time  by  4,  will  reduce 
them  to  minutes  of  longitude  ;  dividing  minutes  of  time  by  4,  will 
reduce  them  to  degrees,  &c.  Hence, 


.  To  find  the  difference  of  time  between  two  places  from 
the  difference  of  their  longitude. 

Reduce  the  difference  of  longitude  to  minutes  ;  multiply  tJiem  by  4, 
and  the  product  will  be  the  difference  of  time  in  seconds,  which 
may  be  reduced  to  hours  and  minutes. 

OBS.  When  the  difference  oflongitude  consists  of  degrees  only,  we  may  mul- 
tiply them  by  4,  and  the  product  will  be  the  answer  in  minutes. 

58.  The  difference  of  longitude  between  New  York  and  Cin- 
cinnati is  10°  26'  :  what  is  the  difference  in  their  time? 

Solution.  —  10°  and  26'  =  G26';  (Art.  281;)  now  626'X4= 
^504  seconds  of  time  ;  and  2504  sec.  -r-60  =  41  min.  44  sec.  Ans. 

59.  The  difference  of  longitude  between  Albany  and  Boston  is 
2°  9'  :  what  is  the  difference  in  their  time  ? 

60.  The  difference  of  longitude  between  Albany  and  Detroit  is 
9°  45'  :  what  is  the  difference  in  their  time  ? 

61.  The  difference  of  longitude  between  New  Haven  and  New 
Orleans  is  17°*10'  :  what  is  the  difference  in  their  time? 

62.  The  difference  of  longitude  between  Charleston,  S.  C.  and 
Mobile  is  8°  27'  :  what  is  the  difference  in  their  time? 

63.  The  difference  of  longitude  between  New  York  and  Canton 
is  187°  3'  :  what  is  the  difference  in  their  time  ? 

293*  To  find  the  difference  of  longitude  between  two  places 
from  the  difference  in  their  time. 

Reduce  the  given  difference  of  time  to  seconds  ;  divide  them  by  4, 
and  the  quotient  will  be  the  difference  of  longitude  in  minutes,  which 
may  be  reduced  to  degrees.  (Art.  281.) 

OBS.  When  there  are  no  seconds  in  the  difference  of  time,  we  may  divide 
the  minutes  by  4,  and  the  quotient  will  be  the  answer  in  degrees. 

Q.CIST,—  292.  How  find  the  difference  of  time  between  two  places  from  their  differ 
ence  of  longitude  ?  293.  How  find  the  difference  of  longitude  from  the  difference  of  time  1 


176  REDUCTION.  [SECT.  VIII. 

64.  A  whip  sailed  from  Boston  to  Liverpool ;  on  the  fourth  day 
the  master  took  an  observation  of  the  sun  at  noon,  and  found  by 
his  chronometer  that  it  was  1  hr.  5  min.  and  40  sec.  earlier  than 
the  Boston  time :  how  many  degrees  east  of  Boston  was  the  ship  ? 

Solution. — 1  hr.  5m.  40  sec.=3940  sec.,  (Art.  281,)  and  3940 
sec.-— 4  =  985'.  The  ship  had  therefore  sailed  985'  east,  which 
i*  equal  to  16°  25'.  Ans. 

65.  The  difference  of  time  between  Albany  and  Buffalo  is  19 
minutes :  what  is  the  difference  of  their  longitude  ? 

66.  The  difference  of  time  between  Richmond  and  New  Orleans 
is  51  min.  4  sec. :  what  is  the  difference  of  their  longitude  ? 

67.  The  difference  of  time  between  Boston  and  Cincinnati  is 
53  min.  32  sec. :  what  is  the  difference  of  their  longitude  ? 

COMPOUND  NUMBERS  REDUCED  TO  FRACTIONS. 

294.  That  one  concrete  number  may  properly  be  said  to  be  a 
part  of  another,  the  two  numbers  must  necessarily  express  objects 
of  the  same  kind,  or  objects  which  can  be  reduced  to  the  same 
kind  or  denomination.  Thus,  1  penny  is  Tj-Jr  of  a  pound,  but  1 
penny  cannot  properly  be  said  to  be  a  part  of  a  foot,  or  of  a  year  ; 
for,  feet  and  years  cannot  be  reduced  to  pence.  So,  1  orange  is  •£ 
of  5  oranges  ;  but  1  orange  cannot  be  said  to  be  -^  of  5  apples,  or 
5  pumpkins;  for  apples  and  pumpkins  cannot  be  reduced  to 
oranges. 

Ex.  1.  Reduce  2s.  Yd.  to  the  fraction  of  a  pound. 

Analysis. — The  object  in  this  example  is  to  find  what  part  of 
1  pound,  2s.  Vd.  is  equal  to.  To  ascertain  this,  we  must  reduce 
both  the  given  numbers  to  the  same  denomination,  viz :  pence. 
Now  2s.  7d.=31d.,  and  £l=240d.  (Art.  281. 1.)  The  question, 
therefore,  resolves  itself  into  this :  what  part  of  240  is  31  ?  The 
answer  is  ^Voj  consequently  2s.  7d.  (3 Id.)  is  -iftV  of  a  pound. 
Hance, 

QUKBT   -294   When  can  one  concrete  number  be  said  to  be  a  part  of  another? 


ARTS.  294,  296.]  REDUCTION.  177 

295.  To  red  a  compound  number  to  a  common  fraction, 
of  a  higher  denomination. 

First  reduce  the  given  compound  number  to  the  lowest  denomina- 
tion  mentioned  for  tJie  numerator  /  then  reduce  a  UNIT  of  tlie  de» 
nomination  of  the  required  fraction  to  the  same  denomination  as  the 
numerator,  and  the  result  will  be  the  denominator.  (Art.  281.) 

OBS.  1  The  given  number,  and  that  of  which  it  is  said  to  be  o.partt  must, 
in  all  cases,  be  reduced  to  the  same  denomination.  (Art.  294.) 

2.  When  the  given  number  contains  but  one  denomination,  it  of  coarse  re- 
quires no  reduction. 

If  the  given  number  contains  a  fraction,  the  denominator  of  the  fraction  is 
the  lowest  denomination  mentioned.  Thus,  in  6|s.,  the  lowest  denomination 
is  fourths  of  a  shilling;  in  -f  far.,  the  lowest  denomination  is  fifths  of  a  farthing 

2.  Reduce  f  of  a  penny  to  the  fraction  of  a  pound. 

Solution.  —  Since  sevenths  of  a  penny  is  the  lowest  and  only 
denomination  given,  we  simply  reduce  «€l  to  sevenths  of  a  penny 
for  the  denominator.  Now  £l  =  240d.,  and  240d.x7=1680. 
Ans.  X-rcVo*  or  <£-T8T'  Hence, 


296.  To  reduce  a  fraction  of  a  lower  denomination  to  an 
equivalent  fraction  of  a  higher  denomination. 

Reduce  a  unit  of  the  denomination  of  the  required  fraction  to 
the  same  denomination  as  the  given  fraction,  and  the  result  will  be 
the  denominator. 

Or,  divide  the  given  fraction  by  the  same  numbers  as  in  reducing 
whole  compound  numbers  to  higher  denominations.  (Art.  281.  II.) 
Thus  in  the  last  example,  fd.-rl2=-&s.,  (Art.  227,)  and  As--r- 
20=£rG-<W>==£ldro.  Ans. 

OBS.  When  factors  common  to  the  numerator  and  denominator  occur  th* 
operation  may  be  shortened  by  canceling  those  factors.  (Art.  221.) 

3.  Reduce  -$•  of  a  penny  to  the  fraction  of  a  pound. 

Solution.  —  By  the  last  article,  -  =  the  answer. 

7X12X20 

By  Cancelation   --  =  -  —  —  —  £  -  •      Ant 
7X12X20     7X12X20,5        420 

QTTKST.—  295.  How  is  a  compound  number  reduced  to  a  common  fraction  ?  296.  How 
Is  a  fraction  of  a  lower  denomination  reduce1  to  the  fraction  of  0  higher  * 


178  REDUCTION.  [Si:CT.  VIII. 


4.  Reduce  4|s.  tc  the  fraction  of  a  pound.     Ans.  ££  £,  or 

5.  Reduce  4s.  Yd.  to  the  fraction  of  a  pound. 

6.  Reduce  9d.  2£  far.  to  the  fraction  of  a  pound. 
7    What  part  of  £l  is  1  of  1  penny  ? 

8.  What  part  of  1  Ib.  Troy  is  7  ounces  ? 

9,  What  part  of  1  Ib.  Troy  is  16  pwts.  3  grs  ? 

10.  What  part  of  1  Ib.  avoirdupois  is  8  oz.  and  12  drams? 

11.  What  part  of  1  ton  is  14  cwt.  and  15  Ibs? 

12.  What  part  of  1  yd.  is  2  ft.  and  4  inches? 

13.  What  part  of  1  mile  is  82i  rods? 

14.  What  part  of  1  acre  is  45£  rods? 

15.  What  part  of  1  square  rod  is  63  square  feet? 

16.  Reduce  £  of  1  qt.  to  the  fraction  of  a  gallon. 

17.  Reduce  7  gallons  to  the  fraction  of  a  hogshead. 

18.  Reduce  •£  of  1  hour  to  the  fraction  of  a  day. 

19.  Reduce  -f  of  1  minute  to  the  fraction  of  an  hour. 

20.  Reduce  -f  of  1  second  to  the  fraction  of  a  week. 

21.  What  part  of  £3,  5s.  6d.  Ifar.  is  £2,  Is.  3d.  ? 

Solution.  —  Reducing  both  numbers  to  farthings,  £3,  5s  6d.  Ifar. 
—  3145  far.,  and  £2,  Is.  3d.  =  1980  far.  (Art.  295.  Obs.'  .)  No* 
1980  is  iff!  of  3145,  which  is  equal  to  fff.  Ans. 


22.  What  part  of  £2  is  7s.  6d.  ? 

23.  What  part  of  £7,  3s.  is  £3  ? 

24.  What  part  of  2  bushels  is  3  pecks  ? 

25.  What  part  of  10  bushels  is  XO  quarts? 

26.  What  part  of  16  rods  is  40  feet? 

27.  What  part  of  3  weeks  is  2  days  and  7  hours  ? 

28.  What  part  of  2  hhds.  10  gals,  is  45  gals.? 

29.  What  part  of  2  tons,  3  cwt.  is  15  cwt.  65  Ibs.  ? 

30.  What  part  of  1  ton  is  *  Ibs.  10  ounces? 

31.  What  part  of  90°  is  lo  15'  30'  ? 

32.  What  part  of  360°  is  45°  15'  10;-  ? 

33.  What  part  of  3  Ibs.  Troy  is  1  Ib.  3  oz.  ? 

34.  What  part  of  25  Ibs.  Troy  is  10  Ibs.  7  oi<  10  pwtst  ? 

35.  What  part  of  1  acre  is  40  rods  ? 
36    What  nart  of  5  acres  is  14-  acres  ? 


ART  297,  REDUCTION.  179 

FRACTIONAL  COMPOUND  NUMBERS 

REDUCED    TO    WHOLE    NUMBERS    OF    LOWER    DENOMINATIONS. 

Ex.  1.  Reduce  -f  of  £1  to  shillings  and  pence. 

Analysis.  —  f  of  ls.=i-  of  5s.  or  -fs.,  consequently  -£  of  20s.  (£l) 
is  20  times  as  much,  and  -fs.  X  20  =-4-^8.  or  12s.  and  f  of  a  shilling. 

reasoning  as  before,  f  of  ld.=i  of  4d.,  or  £d.,  and  %  of  12d. 
(Ir,.)  is  12  times  as  much  ;  but  ^d  X  12=^d.,  or  6d.  Therefore 
£|  —  12s.  6d.  Ans.  Hence, 


*  To  reduce  fractional  compound  numbers  to  whole  num- 
bers of  lower  denominations. 

First  reduce  the  given  numerator  to  the  next  lower  denomination  ; 
then  divide  the  product  by  the  denominator,  and  the  quotient  mil 
be  an  integer  of  the  next  lower  denomination.  (Art.  281.  I.) 

Proceed  in  like  manner  with  the  remainder,  and  the  several  quo* 
tients  will  be  the  whole  numbers  required. 

OBS.  This  operation  is  the  same  in  principle  as  reducing  htgner  denomina 
tions  of  whole  numbers  to  lower.  (Art.  281.  I.)  Whenever  the  fraction  be 
comes  improper,  it  is  reduced  to  a  whole  or  mixed  number.  (Art.  196.) 

2.  Reduce  £  of  £l  to  shillings.  Ans.  16s. 

3.  Reduce  •£  of  £l  to  shillings  and  pence. 

4.  Reduce  f  of  Is.  to  pence  and  farthings. 

5.  Reduce  f  of  1  Ib.  Troy  to  ounces,  &c. 

6.  Reduce  f  of  1  ounce  Troy  to  pennyweights. 

7.  Reduce  f  of  1  Ib.  avoirdupois  to  ounces,  &c. 

8.  Reduce  f  of  1  cwt.  to  pounds,  <fec. 

9.  Reduce  -|  of  1  ton  to  pounds,  <fcc. 

10.  Reduce  i  of  1  yard  to  feet  and  inches. 

11.  Red  ace  f  of  1  rod  to  feet  and  inches. 

12.  Reduce  $  of  1  mile  to  rods,  feet,  &c. 

13.  Reduce  -f  of  1  gallon  wine  measure  to  quarts,  &c. 

14.  Reduce  1  of  1  hogshead  wine  measure  to  gallons,  &c. 

15.  Reduce  f  of  1  peck  to  quarts,  &c.  Ans.  6  qts.  !•£  pts. 

16.  Reduce  %  of  1  bushel  to  quarts,  <fec. 

17.  Reduce     of  1  hour  to  minutes  and  seconds. 


dug  ST.-  -297.  How  are  fractional  compound  numbers  reduced  to  whole 


180  COMPOC,<D  [SECT.  VIII 

18  R»,duce  -A  of  1  day  to  hours,  &c. 

19.  Reduce  £  of  1  minute  to  seconds. 

20.  Reduce  f  of  1  degree  to  minutes,  &c. 

21.  Reduce  £j-Jr$  to  the  fraction  of  a  penny. 

Solution.  —  We  reduce  the  numerator  to  pence,  the  denomina- 
tion required,  and  divide  it  by  the  denominator,  as  in  the  last 
article.  Thus,  2X20X12=480;  and  480-f-720=tf£.  There- 
fore £jiT>=$n<l.=^=t  or  id.  Ans.  Hence, 


298.  To  reduce  a  fraction  of  a  higher  denomination  to  an 
equivalent  fraction  of  a  lower  denomination. 

Reduce  the  given  numerator  to  the  denomination  of  the  required 
fraction,  and  place  the  result  over  the  given  denominator. 

OBS.  1.  This  process  is  the  same  in  principle  as  to  reduce  a.w7iole  compound 
r  imber  to  a  lower  denomination.    (Art.  281.  I.) 

1   &  When  factors  common  to  the  numerator  and  denominator  occur,  the  op- 
t  \tloti  may  be  shortened  by  canceling  those  factors.     (Art.  221  <) 

2  ^  20  ^  1** 
Thus,  in  the  last  example,  -  —  —  —  -=the  answer, 

2X20X12     2X20  X*2      ., 
By  Cancel,  __=_?_==id.     Ans. 

22.  Reduce  7-}^  of  £1  to  the  fraction  of  a  penny. 

2'3.  Reduce  -^?  of  1  Ib.  avoirdupois  to  the  fraction  of  an  ounce. 

24.  Reduce  g7g64  of  1  mile  to  the  fraction  of  a  rod. 

25.  Reduce  -f^-  of  a  day  to  the  fraction  of  an  hour. 

26.  Reduce  -£j  of  1  week  to  the  fraction  of  1  minute. 

27.  Reduce  £f  of  1  yard  to  the  fraction  of  a  nail. 

28.  Reduce  -^  of  1  bushel  to  the  fraction  of  a  quart. 

29.  Reduce  i9^  of  1  hhd.  wine  measure  to  the  fraction  of  a  quart. 

30.  Reduce  -j2^-  of  1  Ib.  Troy  to  the  fraction  of  an  ounce. 

31.  Reduce  iVn^  of  1  pound  Troy  to  the  fraction  of  a  pwfc. 

32.  Reduce  ^V  of  an  acre  to  the  fraction  of  a  rod. 

33.  Reduce  -rfr  of  a  square  yard  to  the  fraction  of  a  foot. 
84.  Reduce  -g-f^  of  a  degree  to  the  fraction  of  a  second. 

QUEST.—  298.  How  is  a  fraction  of  a  higher  denomination  reduced  to  the  fraction  of  a 
fewer  denomination  1  * 


ARTS.  298-300."]  ADDITION.  181 

ADDITION  OF  COMPOUND   NUMBERS. 

299.  The  process  of  adding  numbers  of  different  dtnomi' 
nations,  is  called  COMPOUND  ADDITION. 

1.  What  is  the  sum  of  £6,  11s.  5d.  1  far. ;  £4,  9s.  6d..2  far. ; 
^3,  12s  8d.  3  far. ;  and  £8,  Cs.  9d.  1  far.  ? 

Operation.  Having  placed  the  farthings  under  fai- 

£      s.      d.  far.  things,  the  pence  under  pence,  &c.,    tt-d 

6    '11    '  5  "  1  add  the    column  of  farthings  together,  as 

4        9   '  6   '  2  in  simple  addition,  and  find  the  sum  is  7, 

3  "  12   '  8     3  which   is   equal  to  Id.    and   3    far.  over. 

8  "    6  "  9  "  1  Set  the  3  far.  under  the  column  of  far- 

23~"    0  "  5  "  3  Ans.    things,  and  carry  the  Id.  to  the  column 

of  pence.     The  sum  of  the  pence  is  29,  which  is  equal  to  2s.  and 

5d.  over.     Place  the  5d.  under  the  column  of  pence,  and  carry 

the  2s.  to  the  column  of  shillings.     The  sum  of  the  shillings  is 

40,  which  is  equal  to  £2,  and  nothing  over.    Write  a  cipher  under 

the  column  of  shillings,  and  carry  the  £2  to  the  column  of  pounds. 

The  sum  of  the  pounds  is  23.     Ans.  £23,  Os.  5d.  3  far. 

3  GO*  Hence,  we  derive  the  following  general 

RULE  FOR  ADDING  COMPOUND  NUMBERS. 

I.  Write  the  numbers  so  that  the  same  denominations  shall  staitd 
under  each  other. 

II.  Beginning  with  the  lowest  denomination,  find  the  sum  yfeach 
column  separately,  and  divide  it  by  that  number  which  it  requires 
of  the  column  added,  to  make  ONE  of  the  next  higher  denomination. 
Set  the  remainder  under  the  column  added,  and  carry  the  quotum  t 
to  the  next  column. 

III.  Proceed  in  this  manner  with  all  the  other  denominations 
except  the  highest,  whose  entire  sum  is  set  down. 

PROOF. —  The  proof  is  the  same  as  in  Simple  Addition.  (Art.  55. 

OBS.  1.  Fractional  compound  numbers  should  be  reduced  to  whole  num 
bers  of  lower  denominations,  then  added  as  above.  (Art.  1GG.) 

QUEST. — 299.  What  is  Compound  Addition  7  300.  How  do  you  write  compound  numbers 
for  addition  ?  Which  denomination  do  you  add  first"?  When  the  sum  of  any  coluiun  it 
found,  what  is  to  be  dene  with  it  1  What  is  done  with  the  last  column  ? 


182  COMPOUND  [SECT.  VIII. 

2.  Compound  Addi  ion  is  the  same  in  principle  as  Simple  Addition. 
In  the  latter,  it  is  true,  we  uniformly  carry  the  tens,  and  in  the  former  we  carry 
for  different  numbers ;  yet  in  each  we  always  carry  for  that  number  which  it 
takes  of  the  order  or  denomination  we  are  adding  to  make  one  in  the  ne» 
higher  order  or  denomination. 

2.  3.  4. 

£      s.      d.  £       s.  d.  £  s.     d.    far. 

16     8       9  25  17  11  68  17     10     3 

856  30  12  10  10  9       6     0 

25     6       8  13  15  7  43  ID     11     2 

50     0      11  Ans.  35  16  9  65  14       81 

5.  A  farmer  sold  to  one  customer  3  tons,  5  cwt.  17  Ibs.  13  oz 
of  hay;  to  another,  4  tons,  7  cwt.  35  Ibs.  12  oz. ;  to  another, 

1  ton,  15  cwt.  63  Ibs.  7  oz. :  how  much  hay  did  he  sell  to  all  ? 

6.  What  is  the  sum  of  15  tons,  6  cwt.  45  Ibs.  5  oz. ;  3  tons, 
17  cwt.  80  Ibs.  6  oz. ;  26  tons,  31  Ibs.  7  oz.  V 

7.  What  is  the  sum  of  21  Ibs.  7  oz.  12  pwts.  10  grs. ;  28  Ibs. 
5  oz.  8  pwts.  7  grs. ;  7  Ibs.  6  pwts.  15  grs. ;  41  Ibs.  6  oz.  20  grs. ; 
9  Ibs.  7  grs.  ? 

8.  What  is  the  sum  of  16  Ibs.  3  oz.  6  pwts.  19  grs. ;  100  Ibs. 
8  oz.  16  pwts. ;  97  Ibs.  5  oz.  10  grs. ;  115  Ibs.  9  oz.  ? 

9.  Add  together  19  rods,  12  ft.  8  in.;  64  rods,  13  ft.  3  in.; 
28  rods,  10  ft.  5  in. ;  60  rods,  9  ft.  11  in. 

10.  Add  together  5  leagues,  2  m.  4  fur.  7  rods,  4  yds. ;  18 
leagues,  2  m.  3  fur.  21  rods,  3  yds. ;  85  leagues,  6  fur.  10  rods, 

2  yds.  1  ft. 

11.  Add  together  19  yds.  3  qrs.  3  na. ;  21  yds.  2  qrs.  1  na. , 
42  yds.  1  qr.  2  na. ;  30  yds.  3  qrs.  2  na. 

12.  Add   together  65  yds.  3  qrs.  1  na. ;  81  yds.  2  qrs.  2  na; 
100  yds.  3  qrs.  1  na. ;  95  yds.  1  qr.  1  na. ;  15  yds.  3  na. ;  28 
yds.  2  qrs. 

13.  Add  together  17  A.  25  r.  29  sq.  ft. ;  49  A.  15  r   4  sq.  ft; 
62  A.  29  r.  31  sq.  ft. ;  10  A.  45  r.  16  sq.  ft. 

14.  Add  together  100  A.  3  R.  12  r. ;  115  A.  2  R.  20  r. ;  160 
A.  1  R.  15  r. ;  91  A.  2  R.  26  r. 

QUEST.— Oi ».  Does  Compound  Addition  differ  from  simple  Addition  1 


ART.  301. J  SUBTRACTION.  183 

15.  One  room  in  a  Louse  contains  15  sq.  yds.  5  ft.  7  in.  of  plas- 
tering; another  10  yds.  7  ft.  30  in. ;  another  9  yds.  6  ft.  25  in.; 
another  7  yds.  5  ft.  63  in.  :  how  much  plastering  is  there  in  all 
of  them  ? 

16.  A  merchant  bought  one  cask  of  oil  containing  73  gals.  3 
qts. ;  another  60  gals.  2  qts. ;  another  40  gals.  1  qt. ;  another  65 
gals.  2  qts. :  how  much  oil  did  he  buy  ? 

17.  What  is  the  sum  of  20  hhds.  41  gals.  3  qts,  2  pts.  3  gi. ; 
81  hhds   20  gals.  1  qt.  1  pt.  3  gi. ;  48  hhds.  19  gals.   2  qts.  1  pt. 

2  gi. ;  81  hhds.  40  gals.  1  gi.  ? 

18.  What  is  the  sum  of  10  wks.  5  d,  12  hrs.  40  min. ;  21  wks. 

3  d.  9  hrs.  15  min.  ;  40  wks.  4  d.  17  hrs.  30  min.  ;  42  wks.  Id.? 

19.  What  is  the  sum  of  40  bu.  3$  pks.  4  qts.  ;  63  bu.  2-J-  pks. 

5  qts. ;  80  bu.  7|  pks.  1  qt. ;  45  bu.  2  pks.  3  qts. ;  90  bu.  1  pk.  ? 

20.  What  is  the  sum  of  7  qrs.  6  bu.  1  pk.  3  qts. ;  27  qrs. 

6  bu.  6  qts.  ;  34  qrs.  1  bu.  6  qts. ;  65  qrs.  6  bu.  3  qts.  ? 

SUBTRACTION   OF   COMPOUND   NUMBERS. 

3O1*    The  process  of  finding  the  difference  between  numbers  of 
different  denominations,  is  called  COMPOUND  SUBTRACTION. 

1.  From  £35,  17s.  6d.  3  far.,  subtract  £16,  9s.  8d.  2  far. 

Operation.  Having  placed  the  less  number  under 

£       s.        d.   far.         the   greater,   with    farthings  under  far- 

35  "  17  "     6  "  3  things,  pence  under  pence,  <fec.,  we  sub- 

16  "    9  '      8  "  2  tract   2    far.    from    3    far.,   and   set  the 

19  "    7  ''  10  "  1  Am.     remainder  1   far.  under  the   column  of 

farthings.     But  8d.  cannot  be  taken  from 

6d. ;  we  therefore  borrow  1  from  the  next  higher  denomination, 
which  is  shillings;  and  Is.  or  12d.  added  to  the  6d.  make  18d. 
Now  8d.  from  18d.  leaves  lOd.  Since  we  borrowed,  we  must 
carry  1  to  the  next  denomination  in  the  lower  number,  as  in  sim- 
ple subtraction.  (Art.  72.)  1  added  to  9  makes  10;  and  10  from 
17,  leaves  7.  Finally,  16  from  35,  leaves  19. 

Ans.  £19,  7s.  lOd.  1  fai. 


QTTKBT.— 301.  What  is  Compound  Subtraction? 


18  i  COMPOUND  [SECT.  VIII 

302»  Hence,  we  derive  the  following  general 

RULE   FOR   SUBTRACTING  COMPOUND  NUMBERS. 

1.  Write  the  less  number  under  the  greater,  so  thai  the  same  de- 
nominations may  stand  under  each  other. 

II.  Beginning  with  the  lowest  denomination,  subtract  the  num- 
ber in  each  denomination  of  the  lower  Line  from  tlie  number  ibove 

't,  and  set  the  remainder  below. 

III.  When  a  number  in  any  denomination  of  the  lower  line  is 
larger  than  the  number  above  it,  borrow  one  of  tJie  next  higher  de- 
nomination and  add  it  to  t/te  number  in  the  upper  line.     Subtract 
as  before,  and  carry  1  to  the  next  denomination  in  tlie  lower  line,  as 
in  subtraction  of  simple  numbers.     (Art.  72.) 

PROOF. —  The  proof  is  the  same  as  in  Simple  Subtraction. 

OBS.  1.  Fractional  compound  numbers  should  be  reduced  to  whole  numbers 
of  lower  denominations,  then  subtracted  as  above.  (Art.  IGG.) 

2.  Compound  Subtraction  is  the  same  in  principle  as  Simple  Subtrac- 
tion In  both  cases,  when  the  number  in  the  lower  line  is  larger  than  that 
above  it,  we  borrow  as  many  units  as  it  takes  of  the  order  or  denomination  we 
are  subtracting  to  make  one  of  the  next  higher  order  or  denomination,  and  in 
both,  we  carry  1  to  the  next  figure  in  the  lower  number. 

2.  From  £48,  17s.  6d.  2  far.,  take  £39,  14s.  9d.  3  far. 

3.  From  £l60i,  6|s.  3|d.,  take  £100|,  8s. 

4.  From  £1000,  take  £500,  6s.  7d.  2  far. 

5.  From  16  cwt.  3  qrs.  15  Ibs.,  take  8  cwt.  2  qrs.  8  Ibs.  6  oz. 

6.  From  85  tons  16  cwt.  39  Ibs.,  take  61  tons  14  cwt.  68  Ibs. 

7.  Subtract  69  m.  41  r.  12  ft.  from  89  m.  10  r.  14  ft. 

8.  Subtract  17  1.  2  m.  3  fur.  4  r.  4  ft.  from  191.  1  m.  2  fur.  15  r. 

9.  Subtract  49  bu.  3  pks.  6  qts.  from  85  bu.  2  pks.  4  qts. 

10.  Subtract  95  qrs.  4  bu.  3  pks.  from  115  qrs.  3  bu.  1  pk. 

11.  Subtract  29  yds.  2  qrs.  3  na.  from  85  yds.  1  qr.  2  na. 

12.  Subtract  55  yds.  2  qrs.  1  na.  from  100  yds. 

13.  Subtract  75  gals.  3  qts.  1  pt.  from  82  gals.  2  qts. 

QUEST. — 302.  How  do  you  write  compound  numbers  for  subtraction  ?  Where  oegm  to 
subtract?  When  the  number  in  the  lower  line  is  larger  than  that  above  it,  what  is  to  be 
doue  ?  Obs.  Does  Compound  Subtraction  differ  from  Simple  Subtraction  ? 


ARTS.  302,  303.]  SUBTRACTION.  185 

15.  A  man  having  140  A.  17  r.  of  land,  sold  5 4  A.  58  r. :  how 
much  had  he  left  ? 

16.  Two  men  having  bought  465  A.  48  r.  of  land,  one  of  them 
wished  to  take  '230  A. :  how  much  would  the  other  have? 

17.  A  farmer  having  144  cords.  55  ft.  of  wood,  sold  87  c.  93  ft. : 
how  much  had  he  left  ? 

18.  In  a  certain  village  there  are  two  public  cisterns  ;  one  con- 
.,ains  446  cu.  ft.  69  in.,  the  other  785  cu.  ft.  95  in. :  what  is  the 
lifference  in  their  capacity  ? 

19.  The  latitude  of  the  Cape  of  Good  Hope  is  30°  55'  15' 
and  that  of  Cape  Horn,  55°  58'  30" :  what  is  their  difference  ? 

20.  The  latitude  of  the  Straits  of  Gibraltar  is  36°  6'  30",  and 
*hat  of  the  North  Cape,  71°  10' :  required  their  difference. 

21.  The  longitude  of  New  York  is  74°  I',  and  that  of  Cincin 
jiati  84°  27' :  required  their  difference. 

22.  From  160  yrs.  11  mo.  2  wks.  5  ds.  16  hrs.  30  min.  40  sec., 
take  106  yrs.  8  mo.  3  wks.  6  ds.  13  hrs.  45  min.  34  sec. 

23.  What  is  the  time  from  Feb.  22d,  1845,  to  May  21st,  1847  ? 

Operation.  May  is  the  5th  month,  and  Feb.  the  2d. 

yr.     mo.   d.  Since  22  days  cannot  be  taken  from  21  d.,  we 

/847  "  5  "  21  borrow  1  mo.  or  30  d. ;  then  say  22  from  51 

1845  "  2  "  22  leaves  29.     1  to  carry  to  2  makes  3,  and  3 

(Lns.  2  "  2  "  29  from  5  leaves  2.    5  from  7  leaves  2.    Hence, 

3O3.  To  find  the  time  between  two  dates. 

Write  the  earlier  date  under  the  later,  placing  the  years  on  the 
left,  the  number  of  the  month  next,  and  tJie  day  of  the  month  on  the 
right,  and  subtract  as  before.  (Art.  302.) 

OBS.  1.  The  number  of  the  month  is  easily  determined  )y  reckoning  from 
January,  the  1st  month,  February  the  2d,  &c.  (Art.  264.) 

2.  In  finding  the  time  between  two  dates,  and  in  casting  interest,  30  day§ 
are  considered  a  month,  and  12  months  a  year. 

3.  Instead  of  setting  down  the  ordinal  number  of  the  month,  as  in  the 
solution  above,  some  prefer  to  write  the  number  of  whole  months  that  have 

QUEST.— 303.  How  do  you  find  the  time  between  two  dates  ?  Obs  In  finding  tima  be 
tween  two  dates,  and  in  casting  interest,  how  many  days  are  considered  a  month  "  **«« 
many  months  a  year  ? 


186  COMPOUND  [SECT.  VIIJ 


in  the  given  year.  E.  g.,  if  the  date  is  Feb.  22d,  1845,  they  would 
write  1  in  the  place  of  months;  because,  it  is  said,  2  whole  months  have  not 
elapsed  in  the  year  1845.  But  it  may  be  doubted  whether  this  method  would 
not  ictid  to  frequent  mistakes. 

Besides,  it  may  be  urged  with  equal  reason,  that  1  ought  to  be  deducted 
from  the  day  of  the  month,  and  1  from  the  year  ;  for  neither  22  whole  days,  nor 
1845  whole  years  had  elapsed  at  the  time  of  the  date,  but  the  22d  day  and  the 
1845th  year  were  then  passing.  In  this  way.  the  subject,  which  in  itself  is 
•iniple,  becomes  intricate  and  perplexing. 

24.  General  Washington  was  born  Feb.  22d,  1732,  and  died 
Dec.  14th,  1799  :  how  old  was  he  ? 

25.  The  Independence  of  the  United  States  was  declared,  July 
4th,  1776  :  how  long  is  it  since? 

26.  A  note  was  given  Aug.  25th,  1840,  and  paid  Feb.  6th, 
1842  :  how  long  did  it  run? 

27.  The  United  States  Exploring  Expedition  sailed  from  Norfolk 
on  the  18th  of  Aug.,  1838,  and  returned  to  New  York  on  the  10th 
of  June,  1842  :  how  long  was  the  voyage  ? 

COMPOUND   MULTIPLICATION. 

3O4«  The  process  of  multiplying  numbers  of  different  denom- 
inations, is  called  COMPOUND  MULTIPLICATION. 

Ex.  1.  What  will  6  cows  cost,  at  £5,  2s.  7|d.  apiece? 

Analysis.  —  Since  1   cow  costs  £5,  Operation. 

2s.  7f  d.,  6  cows  will  cost  6  times  as          £        s.       d.   far. 
much.     Beginning    with   the    lowest          5   "   2  "   7   "  3 
denomination,  6  times  3  far.  are  1  8  far.,  6 

equal  to  4d.  and  2  far.  over.  Set  the  30  "  15  "  10  "  2  Ans. 
2  far.  under  the  denomination  multi- 

plied and  carry  the  4d.  to  the  next  product.  6  times  7d.  are  42d. 
and  4d.  make  46d.,  equal  to  3s.  and  lOd.  Set  the  lOd.  under  the 
pence,  and  carry  the  3s.  to  the  next  product.  6  times  2s.  are  12s 
and  3s.  make  15s.  As  the  product  15s.  does  not  make  one  in  the 
next  denomination,  we  set  it  under  the  column  multiplied.  Fi- 
nally, 6  times  £5  are  £30.  The  answer  is  £30,  15s.  lO^d. 

QUEST.—  304.  What  is  Compound  Multiplication  1 


ARTS.  304,  305.]  MULTIPLI  ATIO*.  IS/ 

3O5*  Hence,  we  deduce  the  following  general 

RULE    FOR   MULTIPLYING    COMPOUND   NUMBERS. 

Multiply  each  denomination  separately,  beginning  with  tlie  loivest, 
and  divide  each  product  by  that  number  which  it  takes  of  the  denom- 
ination multiplied;  to  make  ONE  of  the  next  higher ;  set  down  the 
Wiainder,  and  carry  the  quotient  to  the  next  product,  as  in  addition 
^  compound  numbers.  (Art  300.) 

OBS.  1.  When  the  multiplier  is  a  composite  number,  ft  is  advisable  to  multi- 
ply first  by  one  factor  and  that  product  by  the  other.  (Art.  97.) 

2.  Compound  Multiplication  is  the  same  in  principle  as  Simple  Multiplica- 
tion. In  each  we  carry  for  that  number  which  it  takes  of  the  order  or  denom- 
ination we  are  multiplying,  to  make  one  of  the  next  higher  order  or  denorui 
nation. 

2.  What  will  28  horses  cost,  at  £21,  3s,  7|d.  apiece? 

Operation. 
£        s.      d.    far. 
21    "  3  "   7   "  1  We  multiply  by  the  factors  of  28, 

7  which  are  7  and  4,  and  set  down  each 

148  "  5  "  2  "  3  result  as  above. 

4 


593  "  0  "  11  "  0  Ans. 

3.  What  cost  7  acres  of  land,  at  £35,  6s.  7d.  per  acre? 

4.  What  cost  18  barrels  of  flour,  at  £l,  6s.  8£d.  per  barrel? 

5.  A  man  bought  15  loads  of  hay,  each  weighing  1  T.  270|lbe, : 
what  was  the  weight  of  the  whole  ? 

6.  Multiply  16  tons,  3  cwt.  10f  Ibs.  by  25. 

7.  Multiply  12  Ibs.  3  oz.  16  pwts.  by  56. 

8.  If  1  dollar  weighs  17  pwts.  4£  grs.,  how  much  will  96  dol- 
lars weigh  ? 

9.  Multiply  48  hhds.  15  gals.  2  qts.  1  pt.  by  63. 
10.  Multiply  56  pipes,  1  hhd.  23  gals,  by  100. 

QUEST. — 305.  Where  do  you  begin  to  multiply  a  compound  number?    What  is  dovo 
with  each  product?     Obs.  When  the  multiplier  is  a  compos'jte  number,  how 
Does  it  differ  from  Simple  Multiplication  ? 
T.H. 


188  COMPOUND  [SECT.  VIII. 

11.  Bought  72  pieces  of  cl:th,  each  containing  32f  yards: 
how  much  did  they  all  contain  ? 

12.  If  1   cloak  requires   10  yds.   3  qrs.,  how  much  will  500 
cloaks  require  ? 

13.  Multiply  175  miles,  7  fur.  18  rods  by  84. 

14.  Multiply  40  leagues,  2  m.  5  fur.  15  r.  by  50. 

15.  Multiply  149  bu.  12  qts.  by  60. 

16.  Multiply  26  qrs.  7  bu.  3  pks.  5  qis.  by  110. 

17.  Multiply  150  acres,  65  rods  by  52. 

18.  Multiply  310  acres,  3  roods,  3  rods  by  81. 

19.  Multiply  265  cu.  ft.  10  in.  by  93. 

20.  Multiply  148  cords,  29-r3- ft.  by  650. 

21.  Multiply  365  d.  5  hrs.  48  min.  48  sec.  by  35. 
-22.  Multiply  70  yrs.  6  mo.  3  wks.  5  d.  by  17. 

23.  Multiply  75°  40'  21"  by  210. 

24.  If  a  ship  sails  3°  24'  10"  per  day,  how  far  will  she  sail  hi 
60  days  ? 

25.  If  1  acre  produce  45  bu.  26  qts.,  how  much  will  100  acr<  s 
produce  ? 

26.  If  1  barrel  of  flour  requires  4  bu.  3  pks.  5  qts.  of  wheat, 
how  much  will  500  barrels  require  ? 

27.  What  cost  a  chest  of  tea  containing  17  Ibs.,  at  6s.  10£d.  per 
pound  ? 

28.  What  is  the  duty  on  1000  gals,  of  brandy,  at  13s.  7d.  per 
gallon  ? 

29.  What  is  the  duty  on  10560  Ibs.  of  sugar,  at  6d.  3  far.  per 
pound  ? 

30.  What  is  the  duty  on  1500  yards  of  broadcloth,  at  6s.  9|d. 
per  yd.  ? 

31.  If  1  load  of  wood  measures  117  ft.  110  in.,  how  much  will 
40  loads  of  the  same  size  measure  ?  - 

32.  If  1  quarter  of  beef  weighs  216  Ibs.  7  oz.,  how  much  will 
4  quarters  weigh  ? 

33.  If  1  bushel  of  salt  weighs  72  Ibs.  10  oz.,  how  much  will 
850  bushels  weigh  ? 

34.  If  1  cask  of  oil  contains  86  gals.  2  qts.  1  pt.,  ho*  much  wiL 
100  casks  of  the  same  size  contain  ? 


ARTS*  306,  307.]  DIVISION.  189 

COMPOUND   DIVISION. 

3O6»  The  process  of  dividing  numbers  of  different  deramina- 
tions,  is  called  COMPOUND  DIVISION. 

Ex.  1.  Divide  £25,  3s.  4d.  2  far.  by  6. 

Operation.  Beginning  with  the  pounds,  we  find 

£       ?       d    far.  6  is  contained  in  £25,  4  times  and  1 

6)25  "  3  "   4  "  2  over.     Set  the  4  under  the  pounds, 

~T~  3  "  10  "  3  Ans.      and  reduce  the  remainder  £l  to  shil- 
lings, which  added  to  the  3s.  make 

23s.  6  in  23s.  3  times  and  5s.  over.  Set  the  3  under  the  shil- 
lings, and  reduce  the  remainder  5s.  to  pence,  which  added  to  the 
4d.  make  64d.  6  in  64d.,  10  times  and  4d.  over.  Set  the  10 
under  the  pence,  reduce  the  4d.  to  farthings,  and  divide  as  before. 
Ans.  £4,  3s.  lOd.  3  far. 

3O7.  Hence,  we  deduce  the  following  general 

RULE   FOR  DIVIDING  COMPOUND  NUMBERS. 

Begin  with  the  highest  denomination,  and  divide  each  separately. 
Reduce  the  remainder,  if  any,  to  the  next  lower  denomination,  to 
which  odd  the  number  of  that  denomination  contained  in  the  given 
example,  and  divide  the  sum  as  before.  Proceed  in  this  manner 
through  all  the  denominations. 

OBS.  1.  Each  partial  quotient  will  be  of  the  same  denomination,  as  that  part 
of  the  dividend  from  which  it  arose. 

2.  When  the  divisor  exceeds  12,  and  is  a  composite  number,  it  is  advisable 
to  divide  first  by  one  factor  and  that  quotient  by  the  other.  (Art.  129.)     If  the 
divisor  exceeds  12.  but  is  not  a  composite  number,  long  division  may  be  em- 
ployed. (Art.  120.  II.) 

3.  Compound  Division  is  the  same  in  principle,  as  Simple  Division.     Pre- 
fixing the  remainder  to  the  next  figure  of  the  dividend  in  simple  division,  in 
the  same  as  reducing  it  to  the  next  lower  order  or  denomination,  and  adding 
tne  next  figure  to  it. 

QI-EST  — 306.  What  is  Compound  Division  1  307.  Where  do  you  begin  to  divide  a  com- 
pound number  ?  What  is  done  with  the  remainder?  Obs.  Of  what  denomination  is  each 
partial  quotient?  When  the  divisor  is  a  composite  number,  how  proceed  ?  Does  it  dlflet 
from  Simple  Division  ? 


190  DECIMAL  [SECT*  VIII, 

2.  A  man  wished  to  divide  75  cwt.  2  qrs.  10  Ibs.  of  beef  equally 
among  35  families :  how  much  could  he  give  to  each  ? 

Operation. 

cwt.  qrs.    Ibs.  We  divide  by  the  factors  of  35,  which 

7)75  "  2  "  10  are  7  and  5,  and  set  down  each  result  as 

5)1 0  "  3  "  5~  above. 

2  "  0  "  16  Ans. 

8.  Divide  312  Ibs.  9  oz.  18  pwts.  by  43. 

Ans.  7  Ibs.  3  oz.  6  pwta. 

4.  Divide  410  Ibs.  4  oz.  5  pwts.  6  grs.  by  8. 

5.  Divide  786  bu.  18  qts.  by  25. 

6.  A  farmer  raised  1000  hji.  3  pks.  16  qts.  of  wheat  on  40 
acres :  how  much  was  that  per  acre  ? 

7.  A  man  bought  10  horses  for  £200,  15s. :  how  much  did  he 
give  apiece  ? 

8.  Divide  £87,  10s.  7  id.  by  18. 

9.  A  merchant  tailor  put  216  yds.  3  qrs.  of  cloth  into  20 
cloaks :  how  much  cloth  did  each  cloak  contain  ? 

10.  Divide  500  yds.  3  qrs.  2  na.  by  54. 

11.  A  man  traveled  1000  miles  in  12  days:  at  what  rate  did 
he  travel  per  day  ? 

12.  Divide  1500  m.  2  fur.  30  r.  12  ft.  by  7. 

13.  Divide  120  gals.  3  qts.  1  pt.  by  72. 

14.  Divide  400  hhds.  10  gals.  2  qts.  1  pt.  by  9. 

15.  Divide  365  d.  10  hr.  40  min.  by  15. 

16.  Divide  111  yrs.  20  d    13  hrs.  25  min.  10  sec.  by  11. 

17.  Divide  45°  17'  10"  by  25. 

18.  Divide  65  signs  12°  47'  by  41. 

19.  Divide  164  cords,  30  ft.  by  17. 

20.  Divide  410  cords,  10  ft.  21  in.  by  61. 

21.  If  a  chest  of  tea  weighing  96  pounds  cost  £33,  what  will 
I  pound  cost  ? 

22.  If  the  duty  on  a  pipe  of  wine  is  £50,  Os.  6<1.,  what  is  the 
duty  per  gallon  ? 

23.  If  a  person  spends  £200  a  year,  what  are  his  expenses  per 
day? 


ARTS.  308-3 10.  J  FRACTIONS.  191 

SECTION    IX.- 
DECIMAL    FRACTIONS. 

308.  Fractions  which  decrease  in  a  tenfold  ratio,  or  which  ex- 
press simply  tenths,  hundredt/is,  thousandths,  &c.,  are  called  DECI- 
MAL FRACTIONS. 

They  arise  from  dividing  a  unit  into  ten  equal  parts,  then  di- 
viding each  of  these  parts  into  ten  otlier  equal  parts,  and  so  on. 
Thus,  if  a  unit  is  divided  into  10  equal  parts,  1  of  those  parts  is 
called  a  tenth.  (Art.  1*78.)  If  a  tenth  is  divided  into  10  equal 
parts,  1  of  those  parts  will  be  a  hundredth;  for,  TIo-7-10=ToT. 
If  a  hundredth  is  divided  into  10  equal  parts,  1  of  the  parts  will 
be  a  thousandth  ;  for,  1^-5-7- 10=-niVo»  &c.  (Art.  227.) 

OBS.  Fractions  of  this  class  are  called  decimals,  because  they  regularly  de- 
crease in  a  tenfold  ratio.  (Art.  37.  Obs.  2.) 

Decimal  fractions  are  said  to  have  been  invented  by  Lord  Napier,  in  1602. 

309.  Each  order  of  whole  numbers,  we  have  seen,  increases 
in  value  from  units  towards  the  left  in  a  tenfold  ratio ;  and,  con- 
versely, each  order  must  decrease  from  left  to  right  in  the  same 
ratio,  till  we  come  to  units'  place  again.  (Art.  37.) 

3  1  O.  By  extending  this  scale  of  notation  below  units  towards 
the  right  hand,  it  is  manifest  that  the  first  place  on  the  right  of 
units,  will  be  ten  times  less  in  value  than  units'  place ;  that  the 
second  will  be  ten  times  less  than  the  first  y  the  third  ten  times 
less  than  the  second,  &c. 

Thus  we  have  a  series  of  orders  below  units,  which  decrease  in 
a  tenfold  ratio,  and  exactly  correspond  in  value  with  tenths,  hun- 
dred ths,  thousandths,  &c.  (Art.  308.) 

QUEST.— 308.  What  are  Decimal  Fractions  1  From  what  do  they  arise  ?  Oft*.  Why 
called  decimals?  309.  In  what  manner  do  whole  numbers  Increase  and  decrease  1 
310.  By  extending  this  scale  below  units,  what  would  be  the  value  of  the  first  place  on 
the  right  of  units  1  The  second  1  The  third  ?  With  what  do  these  orders  correspond  in 
value  1 


192  DECIMAL  [SECT.  IX 

311*  Decimal  Fractions  are  commonly  expressed  by  writing 
the  numerator  with  a  point  (  .  )  before  it. 

The  point  placed  before  decimals  is  called  the  Decimal  Point, 
or  Separatrix.  Its  'object  is  to  distinguish  the  fractional  parts 
from  whole  numbers. 

If  the  numerator  does  not  contain  so  many  figures  as  there  are 
ciphers  in  the  denominator,  the  deficiency  must  be  supplied  by 
prefixing  ciphers  to  it.  For  example,  -rV  is  written  thus  .1  ;  & 
thus  .2 ;  T^  thus  .3  ;  &c.  T^  is  written  thus  .01,  putting  the 
1  in  hundredths'  place ;  -rfo  thus  .05  ;  &c.  That  is,  tenths  are 
written  in  the  first  place  on  the  right  of  units ;  hundredths  in  the 
second  place ;  thousandths  in  the  third  place,  &c. 

312*  The  denominator  of  a  decimal  fraction  is  always  1  with 
as  many  cipliers  annexed  to  it,  as  there  are  figures  in  the  given  nu- 
merator. (Art.  308.) 

313*  The  names  of  the  different  orders  of  decimals,  or  places 
below  units,  may  be  easily  learned  from  the  following 

DECIMAL   TABLE. 


.s       i 

L ! !  I !  1 1 !  1 1 

.S-c*3     3    -5    ^3    .3     E    "3     S-J= 


3.267145986274 

314*  It  will  be  seen  from  this  table  that  the  value  of  each 
figure  in  decimals,  as  well  as  in  whole  numbers,  depends  upon  the 
place  it  occupies,  reckoning  from  units.  Thus,  if  a  figure  stands 
in  the  first  place  on  the  right  of  units,  it  expresses  tentJis  ;  if  in 

Q.ITE9T. — 311.  How  are  decimal  fractions  expressed  ?  What  is  the  point  placed  before 
decimals  called?  312.  What  is  the  denominator  of  a  decimal  traction?  313.  Repeat 
the  Decimal  Table,  beginning  units,  tenths,  &c.  314.  Upon  what  does  the  value  of  a  de- 
cimal depend  1 


ARTS.  311-316.]  FRACTIONS.  193 

the  second,  hundredths,  &c. ;  each  successive  place  or  order  to- 
wards the  right,  decreasing  in  value  in  a  tenfold  ratio.     Hence, 

315*  Each  removal  of  a  decimal  figure  one  place  from  units 
towards  the  right,  diminishes  its  value  ten  times. 

Prefixing  a  cipher,  therefore,  to  a  decimal  diminishes  its  value 
ten  times  ;  for,  it  removes  the  decimal  one  place  farther  from  units* 
place.  Thus,  .4=^;  but  .04=-r£-o  ;  and  .004=!-^,  &c. ;  for 
ttie  denominator  to  a  decimal  fraction  is  1  with  as  many  ciphers 
annexe!  to  it,  as  there  are  figures  in  the  numerator.  (Art.  312.) 
Annexing  ciphers  to  decimals  does  not  alter  their  value ;  for, 
each  significant  figure  continues  to  occupy  the  same  place  from 
units  as  before.  Thus,  .5=^;  so  .50 =-£&-,  or  -flr,  by  dividing  the 
numerator  and  denominator  by  10;  (Art.  191,)  and  ,500=i\-,/V, 
or  -ft,  &c. 

OBS.  1.  It  should  be  remembered  that  the  units'  place  is  always  the  right 
hand  place  of  a  whole  number.  The  effect  of  annexing  and  prefixing  ciphers 
to  decimals,  it  will  be  perceived,  is  the  reverse  of  annexing  and  prefixing  them 
U>  whole  numbers.  (Art.  98.) 

2.  A  whole  number  and  a  decimal,  written  together,  is  called  a  mixed  num- 
ber. (Art.  183.) 

316.  To  read  decimal  fractions. 

Beginning  at  the  left  hand,  read  the  figures  as  if  they  were  whole 
numbers,  and  to  the  last  one  add  tlie  name  of  its  order.  Thus, 

.7  is  read  7  tenths. 

.36  "  "  36  hundredths. 

.475  "  "  475  thousandths. 

.6342  "  "  6342  ten  thousandths. 

.57834  "  "  57834  hundred  thousandths. 

.284648  "  "  284648  millionths. 

.8913629  "  "  8913629  ten  millionths. 

OBS.  In  reading  decimals  as  well  as  whole  numbers,  the  units'  place  should 
always  be  made  the  starting  point.  It  is  advisable  for  the  learner  to  apply  to 

drKST. — 315.  What  is  the  effect  of  removing  a  decimal  one  place  towards  the  right  1 
What  then  is  the  effect  of  prefixing  ciphers  to  decimals  1  What,  of  annexing  them  1 
Obs.  Which  is  the  units'  place  ?  What  is  a  whole  number  and  a  decimal  written  to« 
gethcr,  called  ?  316.  How  are  decimals  read  1  Obs.  In  reading  decimals,  wha  shook  M 
made  the  starting  point  1 


194  DECIMAL  [SECT.  IX. 

every  figure  the  name  of  its  order,  or  the  place  which  it  occupies,  before  at- 
tempting to  read  them.  Beginning  at  the  units'  place,  he  should  proceed  tow- 
ards the  right,  thus — units,  tenths,  hundredths,  thousandUis,  &c.,  pointing  tc 
each  figure  as  he  pronounces  the  name  of  its  Drder.  In  this  way  he  will  be 
able  to  read  decimals  with  as  much  ease  as  he  can  whole  numbers. 

Read  the  following  numbers : 

(1.)          (2.)          (3.)  (4.) 

.32         .46274       42.008        2.403126 


.03687  17.401  6.004534 

.3624  .00368  23.07  1.100492 

.82344  .00046  81.4389  9.000028 

.13236  .00009  90.0104  8.001249 

(5,)         (6.)          (7.)          (8.) 

12.683  6.00754  4.306702  9.2000076 

20.064  3.0468  0.007006  8.0403842 

35.0072  2.306843  1.13004  0.0000008 

67.4008  1.710386  9.203167  4.3008004 

Note. — Sometimes  we  pronounce  the  word  decimal  when  we  come  to  tho 
separatrix,  and  then  read  the  figures  as  if  they  were  whole  numbers ;  or, 
simply  repeat  them  one  after  another.  Thus,  125.427  is  read,  one  hundred 
twenty-five,  decimal  four  hundred  twenty-seven ;  or,  one  hundred  twenty-fiv*,, 
decimal  four,  two,  seven. 

Write  the  fractional  part  of  the  following  numbers  in  decimals  • 
(9.)  (10.)  (11.)  (12.) 

25^  4r3hr 

30  Mr  6Tffo 

72-n/o  7-i  08  o  41 i H  J  o  9 

13.  Write  9  tenths ;  25  hundredths ;  45  thousandths. 

14.  Write  6  hundredths ;  7  thousandths  ;  132  ten  thousandth*, 

15.  Write  462  thousandths;  2891  ten  thousandths. 

16.  Write  25  hundred  thousandths ;  25  millionths. 

17.  Write  1637246  ten  millionths;  65  hundred  millionths. 

18.  Write  71  thousandths;  7  millionths. 

19.  Write  23  hundredths;  19  ten  thousandths. 

20.  Write  261  hundred  thousandths;  65  hundredths  ;  121 
lionths;  751  trillionths. 

QUEST.— Note.  What  other  method  of  reading  decimals  Is  ment  oned  * 


ARTS.  317-319.]  FRACTIONS.  195 

317.  Decimal  Fractions,  it  will  be  perceived,  differ  from 
Common  Fractions  both  in  their  origin  and  in  the  manner  of  ex- 
pressing them. 

Common  Fractions  arise  from  dividing  a  unit  into  any  number 
of  equal  parts  ;  consequently,  the  denominator  may  be  any  number 
whatever.  (Art.  182.)  Decimals  arise  from  dividing  a  unit  into 
ten  .qual  parts,  then  subdividing  each  of  those  parts  into  ten  other 
•>qujl  parts,  and  so  on;  .consequently,  the  denominator  is  always 
10,  100,  1000,  &c.  (Arts.  308,  312.) 

Again,  Common  Fractions  are  expressed  by  writing  the  numer- 
ator over  the  denominator  /  Decimals  are  expressed  by  writing 
the  numerator  only,  with  a  point  before  it,  while  the  denominator 
is  understood.  (Arts.  182,  311.) 


Decimals  are  added,  subtracted,  multiplied,  and  divided, 
in  the  same  manner  as  whole  numbers. 

OBS.  The  only  thing  with  which  the  learner  is  likely  to  find  any  difficulty, 
is  pointing  off  the  answer.  To  this  part  of  the  operation  he  should  give  par- 
ticular attention. 


ADDITION  OP   DECIMAL   FRACTIONS. 

319.  Ex.  1.  What  is  the  sum  28.35;  345.329;  568.5;  and 
6.485? 

Operation.  Write  the  units  under  units,  tenths  under 

28.35  tenths,    hundredth*    under    hundredths,    &c. ; 

345.329  then,  beginning  at  the  right  hand  or  lowest 

568.5  order,   proceed  thus:    5    thousandths  and  9 

6.485  thousandths  are  14  thousandths.     Write  the 

948.664     Ans.     4  under  the  column  added,  and  carrying  the  1 
to  the  next  column,  proceed  through  all  the 
orders  in  the  same  manner  as  in  simple  addition.     (Art.  54.)     Fi- 
nally, place  the  decimal  point  in  the  amount  directly  under  that 
in  the  numbers  added. 

QUEST.— 317.  How  do  decimals  differ  from  common  fractions  ?  From  what  do  common 
fractions  arise  ?  From  what  do  decimals  arise  1  How  are  common  fractions  expressed  1 
Row  are  decimals  ? 

9* 


196  ADDITION  OF  [SECT.  IX 

3  2O.  Hence,  we  deduce  the  following  general 

RULE   FOR   ADDITION   OF   DECIMALS. 

Write  he  numbers  so  that  the  same  orders  may  stand  under  each 
other,  placing  units  under  units,  tenths  under  tenths,  hundred ths 
under  hundredths,  t&c.  Begin  at  the  right  Jiand  or  lowest  order, 
and  proceed  in  all  respects  as  in  adding  whole  numbers.  (Art.  54.) 

Fr<M  the  right  Imnd  of  tJie  amount,  point  off  as  many  figures' 
for  decimals  as  are  equal  to  the  greatest  number  of  decimal  places 
in  either  of  the  given  numbers. 

PROOF. — Addition  of  Decimals  is  proved  in  the  same  manner  as 
Simple  Addition.  (Art.  55.) 

Note. — The  decimal  point  in  the  answer  will  always  fall  directly  under  the 
decimal  points  in  the  given  numbers. 

EXAMPLES. 

2.  What  is  the  sum  of  25.7;  8.389;  23.056?    Ans.  5*1.145. 

3.  What  is  the  sum  of  36.258  ;  2.0675  ;  382.45  ;  and  7.3984  ? 

4.  What  is  the  sum  of  32.764;  5.78  ;  16.0037  ;  and  49.3046  ? 

5.  What  is  the  sum  of  1.03041;    6.578034;    2.4178;   and 
4.72103  ? 

6.  Add  together  4.25 ;  6.293;  4.612;  38.07;  2.056;  3.248; 
and  1.62. 

7.  Add  together  35.7603;  47.0076;  129.03;  100.007;  and 
JiO.32. 

8.  Add  together  467.3004;  28.78249;  1.29468;  and  3.78241. 

9.  Add  together  21.6434 ;  800.7  ;  29.461  ;  1.7506  ;  and  3.45. 

10.  Add  together  45.001;    163.4234;    20.3045;    634.2104; 
and  234.90213. 

11.  Add  together  293.0072  ;  89.00301;  29.84567;  924.00369; 
and  72.39602. 

12.  Add  together  1.721341;   8.620047;   51.720345;    2.684; 
and  62.304607. 

13.  Add  together  1.293062;  3.00042  ;  9.7003146;  3.600426; 
7.0040031 ;  and  8.7200489. 

Q.UKST. — 330.  How  are  decimals  added  ?  How  point  off  the  answer  1  How  is  addition 
of  decimals  proved  1 


ARTS.  320-321.]  DECIMAL  FRACTIONS.  197 

14.  Add   together    394.61;    81.928;    3624.8103;    640.203; 
6291.302;  721.004;  and  3920.304. 

15.  Add  together  25  hundredths,  8  tenths,  65  thousandths, 
16  hundredths,  142  thousandths,  and  39  hundredths. 

16.  Add  together  9  tenths,  92  hundredths,  162  thousandths, 
489  thousandths,  and  92  millionths. 

17.  Add  together  45  thousandths,   1752  millionths,  624  ten 
millionths,  and  24368  millionths. 

18.  Add  together  29  hundredths,  7  millionths,  62  thousandths, 
and  12567  ten  millionths. 

19.  Add  together  95  thousandths,  61  millionths,  6  tenths,  11 
hundredths,  and  265  hundred  thousandths. 

20.  Add  together  1   tenth,  2  hundredths,  16   thousandths,  7 
millionths,  26   thousandths,  95   ten  millionths,  and  7  ten  thou- 
sandths. 

21.  Add  together  96  hundred  thousandths,  92  millionths,  25 
hundredths,  45  thousandths,  and  7  tenths. 

22.  Add  together  85  thousandths,  17  hundredths,  36  ten  thou- 
sandths, 58  millionths,  363  hundred  thousandths,  185  millionths, 
and  673  ten  thousandths. 


SUBTRACTION  OF  DECIMAL   FRACTIONS. 

321.  Ex.   1.  From  425.684  subtract  216.96. 

Operation.  Having  written  the  less  number  under  the 

425.684  greater,  so  that  units  may  stand  under  units, 

216.96  tenths  under  tenths,  &c.,  we  proceed  exactly 

208.724.  Ans.     as  in  subtraction  of  whole  numbers.  (Art.  72.) 

Thus  0  thousandths  from  4  thousandths  leaves 

4  thousandths.     Write  the  4  in  the  thousandths'  place.     As  the 

next  figure  in  the  lower  line  is  larger  than  the  one  above  it, 

we  borrow  10.     Now  9  from  16  leaves  7;  sot  the  7  under  th 

column  and  carry  1  to  the  next  figure.  (Art.  72.)     Proceed  in  the 

same  manner  with  the  other  figures  in  the  lower  number.     Finally, 

place  the  decimal  point  in  the  remainder  directly  luider  that  ia 

the  given  number. 


198  SUBTRACTION  OF  [Sfif  T     IX. 

322*  Hence,  we  deduce  the  following  general 

RULE  FOR  SUBTRACTION  OP  DECIMALS. 


Write  the  less  number  under  the  greater,  with  units  under 

under  tenths,  hundredths  under  hundred  ths,  d'c. 
as  in  whole  numbers,  and  point  off  the  answer  as  in  addition  <J/" 
decimals.  (Art.  320.) 

PROOF.  —  Subtraction  of  Decimals  is  proved  in  the  same  manurr 
as  Simple  Subtraction.  (Art.  73.) 

ATr?te.—  When  there  are  blank  places  on  the  right  hand  of  the  upper  num- 
ber, they  may  be  supplied  by  ciphers  without  altering  the  value  of  the  decimal, 
(Art.  315.) 

EXAMPLES. 

2.  From  456.0546  take  364.3123.  Ans.  91.7423. 

3.  From  1460.39  take  32.756218. 

4.  From  21.67  take  .682349. 

5.  From  81.6823401  take  9.163. 

6.  From  100.536  take  19.36723. 

7.  From  .076345  take  .009623478. 

8.  From  1  take  .99. 

0.  From  10  take  .000001. 

10.  From  65.00001  take  .9682347. 

11.  From  24681  take  .87623. 

12.  What  is  the  difference  between  25  and  .25  ? 

13.  What  is  the  difference  between  3.29  and  .999  ? 

14.  What  is  the  difference  between  10  and  .0000001  ? 

15.  What  is  the  difference  between  9  and  9.99999  ? 

16.  What  is  the  difference  between  4636  and  .4654  ? 

17.  What  is  the  difference  between  25.6050  and  567.392  ? 

18.  What  is  the  difference  between  76.2784  and  29.84234? 

19.  What  is  the  difference  between  .0000001  and  .0001  ? 

20.  What  is  the  difference  between  .0000004  and  .00004  ? 

21.  What  is  the  difference  between  32  and  .00032  ? 


Q.UKBT  — 322.  How  are  decimals  subtracted  ?    How  point  off  the  answer  1    Flow  is  t  ib 
traction  of  decimals  proved  ? 


ARTS  322,  323.]      DECIMAL  FRACTIONS.  J99 

22.  \Vhat  is  the  difference  between  .00045  and  45  ? 

23.  What  is  the  difference  between  .00000099  and  99  ? 

24.  From  1  thousandth  take  1  millionth. 

25.  From  7  hundred  take  7  hundredths. 

26.  From  29  thousand  take  92  thousandths. 

27.  From  256  millions  take  256  thousandths. 

28.  From  46  hundredths  take  46  thousandths. 

29.  From  95  thousandths  take  999  ten  thousandths. 

30.  From  1  billionth  take  1  trillion th. 

31.  From  2874  millionths  take  211  billionths. 

32.  From  6231  hundred  thousandths  take  154  millionths. 

33.  From  7213  ten  thousandths  take  431  hundred  thousandths 

34.  From  8436  hundred  millionths  take  426  ten  billionths. 


MULTIPLICATION  OF   DECIMALS. 

323*  Ex.  1.  If  a  man  can  reap  .96  of  an  acre  in  a  day,  how 
much  can  he  reap  in  .5  of  a  day  ? 

Analysis. — Since  he  can  reap  96  hundredths  of  an  acre  in  a 
whole  day,  in  5  tenths  of  a  day  he  can  reap  5  tenths  as  much. 
But  multiplying  by  a  fraction  we  have  seen,  is  taking  a  part  of  the 
multiplicand  as  many  times  as  there  are  like  parts  of  a  unit  in  the 
multiplier.  (Art.  210.)  Hence,  multiplying  by  .5,  which  is  equal 
to  T^  or  £,  is  taking  half  of  the  multiplicand  once.  Now  .96,  or 
^4-2=-^-.  (Art.  227.)  ButTW=.48.  (Art.  311.) 

Operation.          We  multiply  as  in  whole  numbers,  and  pointing 

.96  off  as  many  decimals  in  the  product  as  there  are 

.5  decimal  figures  in  both  factors,  we  have  480,     But 

.480  Ans.      ciphers  placed  on  the  right  of  decimals   do  not 

affect  their  value ;  the  0  may  therefore  be  omitted, 

and  w.e  have  .48  for  the  answer. 

(2.)  (3.)  (4.) 

^Multiply  25.38  360.085  6843.02 

*By                 .42  .0043  6.5 

10.6596"  Ans.  1.5483655  Ans.  44479.630  Ant 


200  MULTIPLICATION    OF  [SECT.  IX 

324.  From  the  preceding  illustrations  we  deduce  the  follow- 
ing general 

RULE  FOR  MULTIPLICATION  OF  DECIMALS. 

Multiply  as  in  whole  numbers,  and  point  off  as  many  figures 
from  tJie  right  of  the  product  for  decimals,  as  there  are  decimal 
places  both  in  the  multiplier  and  multiplicand. 

If  the  product  does  not  contain  so  many  figures  as  there  ar 
dt'timals  in  loth  factors,  supply  the  deficiency  by  prefixing  ciphers. 

PROOF. — Multiplication  of  Decimals  is  proved  in  the  same  man* 
rer  as  Simple  Multiplication. 

OBS.  The  reason  for  pointing  off  as  many  decimal  places  in  the  product  as 
there  are  decimals  in  both  factors,  may  be  illustrated  thus : 

Suppose  it  is  required  to  multiply  .25  by  .5.  Supplying  the  denominators 
.25=-i*\,  and  .5=-^.  (Art.  312.)  Now -^-xA^TWo-  (Art.  215.)  But 
tVuV—125  5  (Art-  311 ;)  that  is,  the  product  of  .25XA  contains  just  as  many, 
decimals  as  the  factors  themselves.  In  like  manner  it  may  be  shown  that  the 
product  of  any  two  or  more  decimal  numbers,  must  contain  as  many  decimal 
figures  as  there  are  places  of  decimals  in  the  given  factors. 

EXAMPLES. 

Ex.  1.  In  1  rod  there  are  16.5  feet:  how  many  feet  are  there 
in  41.3  rods? 

2.  In  1  degree  there  are  69.5  statute  miles :  how  many  miles 
are  there  in  360  degrees  ? 

3.  In  1  barrel  there  are  31.5  gallons:   how  many  gallons  in 
65.25  barrels  ? 

4.  In  1  inch  there  are  2.25  nails :  how  many  nails  are  there  in 
60.5  inches  ? 

5.  In  1  square  rod  there  are  30.25  square  yards :  how  many 
square  yards  are  there  in  26.05  rods  ? 

6.  In  one  square  rod  there  are  272.25  square  feet :  how  many 
square  feet  are  there  in  160  rods  ? 


QUEST. — 324.  How  are  decimals  multiplied  together  1  How  do  you  point  off  the  prod- 
uct? When  the  product  does  not  contain  so  many  figures  as  there  are  decimals  in  both 
(actors,  what  is  to  be  done  ?  How  is  multiplication  of  decimals  proved  ? 


ARTS.  324,  325.]  DECIMALS.  201 

7.  How  many  square  rods  are  there  in  a  field  60.5  rods  long 
wid  40.75  rods  wide? 

Multiply  the  following  decimals  : 

8.  1.0013X.25.  21.  40.4369X1.2904. 

9.  44.046 X. 43.  22.  100.0008 X-000306. 

10.  3.6051X4.1.  23.  75.35060X62.3906. 

11.  0.1003X6.12.  24.  31.50301  X  17.0352. 

12.  8.0004X.004.  25.  0.000713X2.30561. 

13.  35.601X1.032.  26.  42.10062X3.821013. 

14.  213.02X4.318.  27.   1.0142034X0620034. 

15.  0.0006 X. 00012.  28.  25067823 X. 0000001 

16.  0.3005X-0035.  29.  64.301257X1.000402. 

17.  10.2106X38.26.  30.  394.20023 X-00000003. 

18.  164.023X1.678.  31.  2564.21035X4.300506. 

19.  9.40061X15.812.  32.  840003.1709X112.10371. 

20.  7.31042X10.021.  33.  0.834567834X-00000008, 

CONTRACTIONS   IN  MULTIPLICATION  OF  DECIMALS. 
CASE    I. 

325.  When  the  multiplier  is  10,  100,  1000,  &c.,  the  multi- 
plication may  be  performed  by  simply  removing  flip  decimal  point 
as  many  places  towards  the  right,  as  there  are  ciphers  in  the  mul- 
tiplier.  (Arts.  99,  3'24.) 

1.  Multiply  85.4321  by  100.  Ans.  8543.21. 

2.  Multiply  42930.213401  by  10. 

3.  Multiply  1067.2350123  by  100. 

4.  Multiply  608.34017  by  1000. 

5.  Multiply  30.467214067  by  10000. 

6.  Multiply  446.3214032  by  100000. 

7.  Multiply  21.3456782106  by  100000. 

8.  Multiply  5  tenths  by  1000. 

9.  Multiply  75  hundredths  by  100000. 

10.  Multiply  65  ten  thousandths  by  1000 

11.  Multiply  48  hundred  thousandths  by  100000. 

QUEST.— 325.  How  proceed  when  the  multiplier  is  10, 100,  fee.  1 


202 


MULTIPLICATION    OF 


[SECT.  IX. 


12.  Multiply  248  thousandths  by  10000. 

13.  Multiply  381  ten  thousandths  by  10000. 

14.  Multiply  6504  ten  millionths  by  100000. 

15.  Multiply  834  thousandths  by  1000000. 

16.  Multiply  1  millionth  by  10000000. 

CASE    II. 

326.  When  the  number  of  decimal  places  in  the  multiplier 
and  multiplicand  is  large,  the  number  of  decimals  in  the  product 
must  also  be  large.  But  decimals  below  the  fifth  or  sixth  ord^r, 
express  so  small  parts  of  a  unit,  that  when  obtained,  they  »r« 
commonly  rejected.  It  is  therefore  desirable  to  avoid  the  imno- 
cessary  labor  of  obtaining  those  which  are  not  to  be  used. 

17.  It  is  required  to  multiply  1.3569  by  .36742,  and  retain 
five  places  of  decimals. 


First  Operation. 
1.3569 
.36742 


.40707 

8141 

949 

54 


83 

276 

7138 


.49855  2198  Ans. 


It  is  evident  from  the  nature  of  decimal 
notation,  that  if  the  partial  product  of 
each  figure  in  the  multiplier  is  advanced* 
one  place  to  the  right  instead  of  the  left, 
the  operation  will  correspond  with  the  de- 
scending scale,  and  at  the  same  time  will 
give  the  true  product.  (Art.  86.  Obs.  ?.) 
But  since  only  five  decimals  are  required, 
those  on  the  right  of  the  perpendicular 
are  useless.  Our  present  object  i?  tc 
show  how  the  answer  can  be  obtained  without  them. 

Beginning  at  the  right  hand,  we  will  firsf 
multiply  the  multiplicand  by  the  tenths'  figure 
of  the  multiplier,  and  place  the  first  figure  of 
the  partial  product  under  the  figure  multiplied. 
In  obtaining  the  second  partial  product,  (i.  e. 
multiplying  by  6,)  it  is  plain  we  may  omit  the 
right  hand  figure  of  the  multiplican  1,  for  if 
multiplied,  its  product  will  fall  to  the  right  of 
the  perpendicular  line,  and  therefore  will  not 


ARTS.  326,  327.]  DECIMALS.  203 

Le  used.  But  if  we  multiply  9  into  6,  the  product  will  be  54 ; 
consequen  ly  there  would  be  5  to  carry  to  the  next  product ;  we 
therefore  3arry  5  to  36,  which  makes  41.  Again,  in  (hi  third 
partial  product,  (i.  e.  in  multiplying  by  7,)  we  may  omit  the  two 
right  hand  figures  of  the  multiplicand  ;  for,  their  product  will  fall 
to  the  right  of  the  perpendicular  line.  But  by  recurring  to  the 
rejected  figures,  it  will  be  seen  that  the  product  of  7  into  6  is  42, 
and  6  to  carry  make  48 ;  we  therefore  add  5  to  the  product  of 
?  'nto  5,  because  48  is  nearer  50  than  40 ;  consequently  it  is 
nearer  the  truth  to  carry  5  than  to  carry  4.  In  the  fourth  partial 
product  we  may  omit  the  three  right  hand  figures,  and  in  the  fifth 
or  lant,  the  four  light  hand  figures. 

18  Multiply  .235*6  by  .3765,  and  retain  4  decimals  in  the 
product. 

Operation.  Multiplying  as  before,  the  first  figure  of  the 

.2356  partial  product  must  be  set  in  the  fifth  order, 

.3765  or  one  place  to  the  right  of  the  figure  multi- 

.0707  plied  ;  for,  there  are  4  decimals  in  the  multipli- 

165  cand  and  the  one  by  which  we  multiply  makes 

14  5.  (Art.  324.)     But  since  we  wish  to  retain 

1  only  4  decimals  in  the  product,  we  may  omit 

.0887   Ans.        this  figure,  carrying  2  to  the  next  product. 

Proceed  in  the  same  manner  with  the  other 

figures  in  the  multiplier.     Finally,  the  sum  of  the  partial  products 

which  are  retained,  is  the  answer  required.     Hence, 

327.  To  multiply  decimals  and  retain  only  a  given  number 
of  decimal  figures  in  the  product. 

Count  off  in  the  multiplicand  as  many  decimal  places  less  out,, 
as  are  required  in  the  product.  Then  beginning  at  the  right  hand, 
figure  counted  off,  multiply  the  multiplicand  by  tJie  tenths  or  first 
decimal  figure  of  the  multiplier,  and  set  the  first  figure  of  the 
partial  product  one  place  to  the  right  of  the  figure  multiplied,  in- 
creasing it  by  the  nearest  number  of  tens  tJiat  would  arise  from  the 

UUBST.  -327.  How  multiiAj  decimals,  and  retain  a  given  number  of  figures  in  the  product? 


201  MULTIPLICATION    OF  [SECT.   IX 

rejected  j\ (jure  if  multiplied.  Next  multiply  by  the  second  decimal 
figure.,  omitting  the  next  right  hand  figure  of  the  multiplicand  and 
carrying  as  before.  Proceed  in  the  same  manner  with  all  t/M  figures 
of  the  multiplier  whose  product  will  come  under  the  decimal  places 
counted  off,  omitting  an  additional  figure  on  the  right  of  tJie  mul- 
tiplicand, as  you  multiply  by  each  successive  figure,  and  set  the 
first  figure  of  each  partial  product  under  that  of  the  preceding. 
Finally,  from  the  sum  of  the  partial  products,  cut  of  the  required 
nvmber  of  decimals,  and  the  result  will  be  the  answer. 

OBS.  1.  In  order  to  determine  where  to  place  the  decimal  point  in  the  prod- 
uct, we  have  only  to  observe  that  the  product  of  the  right  hand  figure  of  the 
multiplicand  into  the  tenths  of  the  multiplier  is  of  the  order  denoted  by  the  stun 
of  the  orders  of  the  two  figures  multiplied ;  (Art.  324 ;)  and  when  the  multi- 
plier is  tenths  it  is  of  the  order  next  lower  than  the  figure  multiplied.  For  this 
reason  the  first  partial  product  is  set  one  place  to  the  right  of  the  figure  multi- 
plied. But  since  we  count  off  one  decimal  less  than  is  required  in  the  prod- 
uct, the  right  hand  figure  in  the  sum  of  the  partial  products  must  consequently 
be  the  right  hand  decimal  place  in  the  answer. 

2.  If  the  multiplier  contains  units,  lens,  hundreds,  &c.,  in  multiplying  by  the 
units,  we  must  begin  one  figure  to  the  right  of  those  counted  off,  and  set  the 
first  figure  of  the  partial  product  under  the  figure  multiplied.  In  multiplying 
by  the  tens,  we  must  begin  two  figures  to  the  right  of  those  counted  off,  and 
set  the  first  figure  of  the  partial  product  under  that  of  the  units;  in  multiply- 
ing by  the  hundreds,  we  must  begin  three  figures  to  the  right,  and  set  the  first 
figure  of  the  partial  product  under  that  of  the  preceding,  &c.  This  will  bring 
the  same  orders  under  each  other. 

19.  Multiply  .72543414  by  .24826421  retaining  5  decimal 
places  in  the  product. 

Operation. 

.7254'3414  Having  counted  off  4  decimals  in  the  mul- 

.2482  6421  tiplicand,  increase  the  product  of  2   into  4 

1450  9  by  1,  because  the  product  of  the  3  rejected 

290  2  into  2  is  nearer  10  than  0.     Set  the  9  one 

58  0  place  to  the  right  of  the  figure  multiplied. 

1  4  The  4  in  the  last  partial  product,  is  the 

4  number  which  would  be  carried  to  this  order, 

.1800  9  Ans.  if  the  V  were  multiplied  by  6. 


ARTS.  327,  328.]  DECIMALS.  205 

20.  Multiply  67.1498601  by  92.4023553  retaining  fear  deci- 
mals in  the  product. 

Operation. 

67.149'8601  In  this  operation  we  multiply  first  by  the 

92.402  3553         tens  figure  of  the  multiplier,  beginning  two 

6043.487  4  places  to  the  right  of  those  counted  off  in  the 

134  299  7  multiplicand.     It  is  immaterial  as  to  the  re- 

26  859  9  suit  whether  we  multiply  by  the  tenths  first, 

]  34  3  or  by  the  units,  tens,  or  hundreds,  provided 

201  we  set  the  first  figure  of  the  partial  product  in 

3  4  its  proper  place.  (Art.  327.  Obs.  2.) 

3_ 

6204.805  1  Ans. 

21.  Multiply  .863541  by  .10983  retaining  5  decimal  places. 

22.  Multiply  1.123674  by  1.123674  retaining  6  decimal  places. 

23.  Mid  ti  ply  .26736  by. 28758  retaining  4  decimal  places. 

24.  Multiply  .1347866  by  .288793  retaining  7  decimal  places. 

25.  Multiply  .681472  by  .01286  retaining  5  decimal  places. 

26.  Multiply  .053407  by  .047126  retaining  6  decimal  places. 

27.  Multiply  .3857461  by  .0046401  retaining  6  decimal  places. 

DIVISION   OP  DECIMAL   FRACTIONS. 

328*  Ex.  1.  How  many  bushels  of  oats,  at  .2  of  a  dollar  a 
6ushel,  can  you  buy  for  .84  of  a  dollar  ? 

Analysis. — Since  2  tenths  of  a  dollar  will  buy  1  bushel,  84 
hundredths  of  a  dollar  will  buy  as  many  bushels,  as  2  tenths  is 
contained  times  in  84  hundredths.  Now  .84=T8^r ;  and  .2=-^, 
ortW-  (Art.  191.)  And  VW-r-iV^A,  or  4-ft.  But,  (Art.  311,) 
.4120-=4.2,  which  is  the  answer  required. 

Operation. 

.2). 84  We  divide  as  in  whole  numbers,  and  point  off 

4.2  Ans.      one  decimal  figure  in  the  quotient. 

OBS.  The  reason  for  pointing  off  one  decimal  figure  in  the  quotient  may  be 
thus  explained. 

We  have  seen  in  the  multiplication  of  decimals,  that  the  product  has  aa 
<nany  decimal  figures,  as  th<»  multiplier  and  multiplicand.  \A.r  "  VM  Now 


206  DIVISION  OF  [SECT.  IX. 

since  the  dividend  is  equal  to  the  product  of  the  divisor  and  quotient,  (Art.  112,) 
it  follows  that  the  dividend  must  have  as  many  decimals  as  the  divisor  and 
qtwtient  together  ;  consequently,  as  the  dividend  has  Iwo  decimals,  and  the 
divisor  but  one,  we  must  point  off  one  in  the  quotient.  In  like  manner  it  may 
be  shown  universally,  that 

329*   The  quotient  must  have  as  many  decimal  figures,  as  the 
decimal  places  in  the  dividend  exceed  those  in  the  divisor  ;  tJiat  is, 
*he  decimal  places  in  the  divisor  and  quotient  together,    :nu$t   be 
qual  in  number  to  those  in  the  dividend. 

2.  What  is  the  quotient  of  3.775  divided  by  2.5  ?     Ans.  1.51. 

3.  What  is  the  quotient  of  .0072  divided  by  2.4. 

Operation.  Since  the  dividend  has  three  decimals 

2.4).0072(.003  Ans.       more  than  the  divisor,  the  quotient  must 

72  have  three  decimals.     But  as  it  has  but> 

one  figure,  we  prefix  two  ciphers  to  it  to 

make  up  the  deficiency. 


OBS.  It  will  be  noticed  that  3,  the  first  figure  of  the  quotient,  denotes 
sandths;  also  the  product  of  2,  the  units  figure  of  the  divisor,  into  the  first  quo- 
tient figure,  is  written  under  the  thousandths  in  the  dividend.     Hence, 

The  first  figure  of  tJie  quotient  is  of  the  same  order,  as  that 
figure  of  the  dividend  under  which  is  placed  the  product  of  the 
units  of  the  divisor  into  the  first  quotient  figure. 

33O.  From  the  preceding  illustrations  we  deduce  the  follow- 
ing general 

RULE  FOR  DIVISION  OF  DECIMALS. 

Divide  as  in  whole  numbers,  and  point  off  as  many  figures  for 
decimals  in  the  quotient,  as  the  decimal  places  in  the  dividend  exceed 
thfise  in  tlie  divisor.  If  the  quotient  does  not  contain  figures  enouoht 
supply  the  deficiency  by  prefixing  ciphers. 

PROOF.  —  Division  of  Decimals  is  proved  in  the  same  manner  of 
Simple  Division.  (Art.  121.) 

OBS.  1.  When  the  number  of  decimals  in  the  divisor  is  the  same  as  thtt  IB 
the  dividend,  the  quotient  will  be  a  whole  number. 

Q.VCST.—  330.  How  are  decimals  divided  How  point  off  the  quotient  1  How  is  division 
of  decimals  proved  ? 


ARTS.  329,  330.] 


DECIMALS. 


207 


2.  When  there  are  more  decimals  in  the  divisor  than  in  the  dividend,  annex 
as  many  ciphers  to  the  dividend  as  are  necessary  to  make  its  decimal  places 
equal  to  th  )se  in  the  divisor.  The  quotient  thence  arising  will  be  a  whole 
number.  (Obs.  I.) 

3  After  all  the  figures  of  the  dividend  are  divided,  if  there  is  a  remainder, 
ciphers  may  be  annexed  to  it  and  the  invasion  continued  at  pleasure.  The 
cipbsrs  annexed  must  be  regarded  as  decimal  places  belonging  to  the  dividend. 

Note. — 1  For  o  binary  purposes,  it  will  be  sufficiently  exact  to  carry  the 
^aotrnt  *u  three  it  four  places  of  decimals  ;  but  when  great  accuracy  is  re- 
quire •,  ;  must  be  carried  farther. 

2.  W  ^en  there  is  a  remainder,  the  sign  -}->  should  be  annexed  to  the  quo- 
tient to  show  that  it  is  not  complete. 

EXAMPLES. 

4.  How  many  boxes  will  it  require  to  pack  71.5  Ibs.  of  butter, 
if  you  put  5.5  Ibs.  in  a  box  ? 

5.  How  many  suits  of  clothes  will  29.6  yds.  of  cloth  make,  al- 
lowing 3.7  yds.  to  a  suit? 

6.  If  a  man  can  walk  30.25  miles  per  day,  how  long  will  it  take 
him  to  walk  150.75  miles? 

7.  How  many  loads  will  134642.156  Ibs.  of  hay  make,  allowing 
1622.2  Ibs.  for  a  load? 

8.  If  a  team  can  plough  2.3  acres  in  a  day,  how  long  will  it 
take  to  plough  63.75  acres? 

9.  How  many  bales  of  cotton  in  56343.75  Ibs.,  allowing  375  Ibs. 
to  a  bale  ? 

Divide  the  following  decimals : 

•  10.  46.84^-7.9. 

11.  1.658H-.25. 

12.  67234-T-.85. 

13.  4.00334—6.31. 

14.  73.8243  — .061. 

15  0.00033  — .011. 

16  236.041  —  1.75. 

17  60.0001-1.0], 
IF  300.402—12.1. 


4.32067  — .001 


20.  0.00006-.003. 

21.  167342-^.002. 

22.  684234.6-2682. 

23.  0.000045  —  9. 

24.  7.231068— .12. 

25.  26.3845-K125. 

26.  4-T-.00001. 

27.  6-^.0000001. 

28.  0.8  r.OOOOJ02. 

29.  6541.234567-^21. 


Obs.  When  the  number  of  decimai  places  in  the  divisor  is  equal  to  that  in  the 
dividend  -hat  is  the  quotient  ?  When  there  are  more  decimals  in  the  divisor  than  in  the 
mvlaem'  *ow  proceed  1  When  there  is  a  remainder,  what  may  be  done  ? 


208 


DIVISION    OF 


[SECT.  IX 


CONTRACTIONS   IN   DIVISION  OF  DECIMALS. 
CASE     I. 

331.  When  the  divisor  is  10,  100,  1000,  &c.,  the  dirision 
may  be  performed  by  simply  removing  tJie  decimal  point  in  t~e 
dividend  as  many  places  towards  the  left,  as  there  are  ciphers  in 
the  divisor,  and  it  will  be  the  quotient  required.  (Arts.  131,  330.) 

1.  Divide  4672.3  by  100.  Ans.  46.723. 

2.  Divide  0.8  by  10000.  Ans.  0.00008. 

3.  Divide  672345.67  by  10. 

4.  Divide  10342.306  by  100. 

5  Divide  42643.621  by  100000. 

6.  Divide  6723000.45  by  100UOOO. 

*;.  Divide  1.2300456  by  100000. 

8.  Divide  2.0076346  by  1000000. 

CASE    II. 

332.  When  the  divisor  contains  a  large  number  of  decimal 
figures,  the  process  of  dividing  may  be  very  much  abridged. 

9.  It  is  required  to  divide  3.2682  by  2.4736,  and  carry  the 
quotient  to  four  places  of  decimals. 

Common  Method. 

2.4736)3.2682(1.3212 

2  4736 


7946 

7420 

525 

494 

~30 

24 


8 
2~0 

72 

480 

736 


7440 
9472 
7968 

Explanation. — We  perceive  the  first  figure  of  the  ijuotient  Trill 
be  a  whole  number ;  for  the  number  of  decimals  in  the  divisor  is 


Qi  «STi— 331.  When  the  divisor  is  10,  100, 1000,  &c.,  how  may  the  division  be  performed  1 


ARTS.  331-333.  J  DECIMALS.  209 

equal  to  that  of  the  dividend.  (Art.  330.  Obs.  1.)  Now  to  obtain 
the  decimals  required,  instead  of  annexing  a  cipher  to  the  several 
remainders,  which  multiplies  them  respectively  by  10,  (Art.  98,) 
we  may  cut  off  a  figure  on  the  right  of  the  divisor  at  each  division, 
which  is  the  same  as  dividing  it  successively  by  10.  (Art.  130.) 
When  we  multiply  the  divisor  by  3,  the  second  quotient  figure, 
we  carry  2  to  the  product  of  3  into  3,  because  the  product  of  3 
into  6,  the  figure  omitted  in  the  divisor,  is  nearer  20  than  10. 
(Art  327.)  We  carry  on  the  same  principle  to  the  first  figure  of 
each  product  of  the  divisor  into  the  respective  quotient  figures. 
Hence, 

333.  To  divide  decimals,  carrying  the  quotient  to  any  re- 
quired number  of  decimal  places. 

For  the  first  quotient  figure  divide  as  usual ;  then  instead  of 
bringing  down  the  next  figure,  or  annexing  a  cipher  to  the  remain- 
der, cut  off  a  figure  on  the  right  of  the  divisor  at  each  successive 
division,  and  divide  by  the  other  figures.  In  multiplying  the  divisor 
by  the  quotient  figure,  carry  for  the  nearest  number  of  tens  that 
would  arise  from  the  product  of  the  figure  last  cut  off  into  the  fig- 
ure last  placed  in  the  quotient.  (Art.  327.) 

OBS.  1.  The  reason  for  this  contraction  may  be  seen  from  the  principle,  that 
a  tenth  of  the  given  divisor  is  contained  in  a.lenth  of  the  dividend,  just  as  many 
times  as  the  whole  divisor  is  contained  in  the  whole  dividend;  (Art.  145;)  for, 
cutting  off  a  figure  on  the  right  of  the  divisor,  and  omitting  to  annex  a  cipher 
to  the  dividend  or  remainder,  is  dividing  each  by  ten.  (Art.  130.) 

2.  When  the  divisor  has  more  figures  than  the  quotient  is  required  to  have, 
including  the  whole  number  and  decimals,  we  may  take  as  many  on  the  left 
of  the  divisor  as  are  required  in  the  quotient,  and  divide  by  them  as  above. 

3.  If  the  divisor  does  not  contain  so  many  figures  as  are  required  in  the  quo- 
tient, we  must  divide  in  the  usual  way,  until  we  obtain  enough  figures  to  make 
up  this  deficiency,  and  then  begin  the  contraction. 

10.  Divide  .4134  by  .3243,  and  carry  the  quotient  to  four 
places  of  decimals. 

11  Divide  .079085  by  .83497,  and  carry  the  quotient  to  five 
places  of  decimals. 

12.  Divide  2.3748  by  1.4736,  and  carry  the  quotient  to  three 
places  of  decimals. 


210  REDUCTION    OF  [SECT.   IX. 

13.  Divide   .3412  by  8.4736,   and  carry  the  quotient  to  five 
places  of  decimals. 

14.  Divide  1   by  10.473654,  and  carry  the  quotient  to  seven 
places  of  decimals. 

15.  Divide  .4312672143  by  .2134123406,  and  carry  the  quo- 
tient to  four  places  of  decimals. 

16.  Divide  .879454  by  .897,   and  carry  the  quotient  to  six 
places  of  decimals. 

REDUCTION  OP   DECIMALS. 

CASE    I. 

334:*  Decimals  reduced  to  Common  Fractions. 
Ex.  1.  Change  the  decimal  .75  to  a  common  fraction. 

Suggestion. — Supplying  the  denominator,  ,75=-1-VL0-.  (Art.  311.) 
Now  -j2^-  is  expressed  in  the  form  of  a  common*fraction,  and,  as 
such,  may  be  reduced  to  lower  terms,  and  be  treated  in  the  same 
manner  as  any  other  common  fraction.  Thus,  T'J/V =•£-&,  or  -f-. 

335.  Hence,  To  reduce  a  Decimal  to  a  Common  Fraction. 

Erase  the  decimal  point ;  then  write  the  decimal  denominator 
under  the  numerator,  and  it  will  form  a  common  fraction,  which 
may  be  treated  in  the  same  manner  as  other  common  fractions. 

2.  Change  .225  to  a  common  fraction,  and  reduce  it  to  the 
lowest  terms.     Ans.  -f$. 

3.  Reduce  .125  to  a  common  fraction,  &c. 

4.  Reduce  .95  to  a  common  fraction,  &c. 

5.  Reduce  .435  to  a  common  fraction,  <fec. 

6.  Reduce  .575  to  a  common  fraction,  <fcc. 

7.  Reduce  .656  to  a  common  fraction,  <fec. 

8.  Reduce  .204  to  a  common  fraction,  &c. 

9.  Reduce  .075  to  a  common  fraction,  &c, 
10.  Reduce  .012  to  a  common  fraction,  &c. 

"11.  Change  .0025  to  a  common  fraction,  &c. 
12.  Change  .1001  to  a  common  fraction,  &c. 

QUEST. — 335.  How  are  Decimals  reduced  to  Common  Fractions  1 


ARTS.  334-337.]  DECIMALS.  21 1 

13.  Change  .1844  to  a  common  fraction,  &c. 

14.  Change  ,0556  to  a  common  fraction,  &c. 
15    Change  .1216  to  a  common  fraction,  &c. 

16.  Change  .2005  to  a  common  fraction,  &c. 

17.  Change  .0015  to  a  common  fraction,  &c. 

CASE     II. 

336.    Common  Fractions  reduced  to  Decimals. 
Ex.  1 .  Change  f  to  a  decimal. 

Suggestion. — Multiplying  both  terms  by  10,  tlie  fraction  be- 
comes -f^-.  Again,  dividing  both  terms  by  5,  it  becomes  -fa. 
(Art.  191 .)  But  -A=.8,  which  is  the  decimal  required.  (Art.  311.) 

Note. — Since  we  make  no  use  of  the  denominator  10  after  it  is  obtained,  we 
may  omit  the  process  of  getting  it ;  for,  if  we  annex  a  cipher  to  the  numerator 
and  divide  it  by  5,  we  shall  obtain  the  same  result. 

Operation. 

5)4.0  A  decimal  point  is  prefixed  to  the  quotient  to 

.8  Ans.       distinguish  it  from  a  whole  number. 
2.  Reduce  -f  to  a  decimal.  Ans.  0.62*5, 

337*  Hence,  to. reduce  a  Common  Fraction  to  a  Decimal. 

Annex  ciphers  to  the  numerator  and  divide  it  by  the  denominator. 
Point  off  an  many  decimal  figures  in  the  quotient,  as  you  Imve  an- 
nexed ciphers  to  the  numerator. 

OBS.  1.  If  there  are  not  as  many  figures  in  the  quotient  as  you  have  an- 
nexed ciphers  to  the  numerator,  supply  the  deficiency  by  prefixing  ciphers  to 
the  quotient. 

2.  The  reason  of  this  rule  may  be  illustrated  thus.  Annexing  a  cipher  to 
the  numerator  multiplies  the  fraction  by  10.  (Arts.  98,  18*}  )  If,  therefore,  the 
numerator  with  a  cipher  annexed  to  it,  is  divided  '  the  \  nominator,  the  quo- 
tient will  obviously  be  ten  times  too  large.  (Art.  141.)  Hence,  in  order  to  ob- 
tain the  true  quotient,  or  a  decimal  equal  to  the  given  fraction,  the  quotient 
thus  obtained  must  be  divided  by  10,  which  is  done  by  pointing  off'  one  figure. 
(Art.  131.)  Annexing  2  ciphers  to  the  numerator  multiplies  the  fraction  by 
100;  annexing  3  ciphers  by  1000.  (fee.,  consequently,  when  2  ciphers  are  an- 
nexed, the  quotient  will  bv1,  100  times  too  large,  and  must  therefore  be  divided 
bi  100;  when  three  ciphers  are  annexed,  the  quotient  will  be  1000  times  too 
I«'ge,  and  must  be  divided  by  1000,  &c.  (Art.  131.) 


T.II. 


Q.UKST. — 337.  How  are  Common  Fractions  reduced  to  Decimals  ? 

10 


4.  £. 

8.  i. 

12.  |. 

5.  i. 

9.  f  . 

13.  i. 

6.  i. 

10.  f. 

14.  1. 

1.  f- 

11.  f. 

15.  i 

20.  Reduce  -t 

}•  to  a  decimal. 

Ans. 

21.  Reduce  -• 

sVr  to  a  decimal. 

Ans. 

212  REDUCTION    OF  [SECT.   IX, 

3.  Reduce  -HP  to  a  mixed  num^r.  Ans.  1.0625. 

Reduce  the  following  fractions  to  decimals : 

16.  -\. 

17.  |. 

18.  f. 

19.  i. 

.  0.166666,  &c. 
Ans.  OJ  23123123,  &c. 

338.  It  will  be  seen  that  the  last  two  examples  cannot  be 
exactly  reduced  to  decimals ;  for  there  will  continue  to  be  a  re- 
mainder after  each  division,  as  long  as  we  continue  the  operation. 

In  the  20th,  the  remainder  is  always  4  ;  in  the  21st,  after  ob- 
taining three  figures  in  the  quotient,  the  remainder  is  the  same  as 
the  given  numerator,  and  the  next  three  figures  in  the  quotient 
are  the  same  as  the  first  three,  when  the  same  remainder  will  re- 
cur again.  The  same  remainders,  and  consequently  the  same  fig- 
ures in  the  quotient,  will  thus  continue  to  recur,  as  long  as  tha  : 
operation  is  continued. 

339*  Decimals  which  consist  of  the  same  figure  or  set  of  fig- 
ures continually  repeated,  as  in  the  last  two  examples,  are  called 
Periodical  or  Circulating  Decimals  ;  also,  Repeating  Decimals,  or 
Repetends. 

OBS.  When  only  a  single  figure  is  repeated,  it  is  more  accurate  to  call  them 
repeating  decimals ;  but  when  two  or  more  figures  recur  at  regular  intervals, 
they  are  very  properly  called  periodical,  or  circulating  decimals. 

34O»  When  a  common  fraction  can  be  exactly  expressed  by  a 
decimal,  the  decimal  is  said  to  be  terminate,  or  finite  ;  but  when 
it  cannot  be  exactly  expressed  by  a  decimal,  it  is  said  to  be  inter' 
l  linate,  or  infinite. 

OBS.  It  seems  to  be  incongruous  to  call  a  fraction  infinite.  (Art.  180.)  The 
term  infinite,  however,  does  not  refer  to  the  vafaie  of  the  fraction,  but  'o  the 
number  Df  decimal  figures  required  to  express  its  value. 

QricsT.— Obs.  When  there  are  not  so  many  figures  in  the  quotient  as  you  have  annexed 
ciphers,  what  is  to  he  done  1  339.  What  are  periodic*  i  or  circulating  dec  jiials  1  310.  Wh«» 
is  a  decimal  terminate  1  When  mterminate  ? 


AIITS.  338-343.]  DECIMALS.  213 

3  4 1  •  If  the  denominator  of  a  common  fraction  when  reduced 
to  its  lowest  term,  contains  no  prime  factors  but  2  and  5,  its 
equivalent  decimal  will  terminate  ;  on  the  other  hand,  if  it  con- 
tains any  other  prime  factor  besides  2  and  5,  it  will  not  terminate. 

Thus  e\  reduced  to  its  lowest  terms,  becomes  -fa,  and  the  prime 
factors  of  20  are  2,  2,  and  5;  that  is,  2(3=2X2X5.  (Art.  165.) 
We  also  find  that  ^=.05  ;  it  is  therefore  terminate.  Again, 
fs~--^s  5  and  the  prime  factors  of  15  are  3  and  5  ;  that  is,  15  = 
8  K  5  ;  and  -^-=.0666666,  &c. ;  it  is  therefore  interminate.  Hence, 

342.  To  ascertain  whether  a  common  fraction  can  be  exactly 
expressed  by  decimals. 

Reduce  the  given  fraction  to  its  lowest  terms,  and  then  resolve  its 
denominator  into  its  prime  factors.  (Art.  341.) 

OBS.  The  truth  of  this  principle  is  evident  from  the  consideration,  that  an- 
nexing ciphers  to  the  numerator,  multiplies  it  successively  by  10 ;  but  2  and  5 
are  the  prime  factors  of  10,  and  are  the  only  numbers  that  can  divide  it  without 
a  remainder.  (Art.  165.  Obs.  2.)  But  any  number  that  measures  another, 
must  also  measure  its  product  into  any  whole  number;  (Art.  161.  Prop.  14;) 
consequently,  when  the  denominator  contains  no  prime  factors  but  2  and  5, 
the  division  will  terminate ;  but  when  it  contains  other  factors,  the  division  can 
n<jt  terminate. 

22.  Will  5^  produce  a  terminate  or  interminate  decimal? 

23.  Will  -f^-  produce  a  terminate  or  interminate  decimal  ? 

24.  Will  -£-£  produce  a  terminate  or  interminate  decimal  ? 

25.  Will  -§V  produce  a  terminate  or  interminate  decimal? 
20.  Will  2-fir  produce  a  terminate  or  interminate  decimal? 

27.  Will  -3^6-  produce  a  terminate  or  interminate  decimal? 

28.  Will  -§VV  produce  a  terminate,  or  interminate  decimal? 

343*  When  the  decimal  is  terminate,  the  number  of  figures  it 
contains,  must  be  equal  to  the  greatest  number  of  times  that  either 
of  the  prime  factors  2  or  5,  is  repeated  in  its  denominator,  whea 
the  given  fraction  is  reduced  to  its  lowest  terms. 

Osa.  The  truth  of  this  principle  may  be  illustrated  thus :  *  =  .5;  that  is  ie 
decimal  terminates  with  one  place ;  for,  the  denominator  2.  is  taken  only  ^nce 
as  a  factor  in  10,  and  therefore  only  one  cipher  is  required  to  be  annexed  t^the 
numerator  to  reduce  it.  Again,  ^  =  .25,  which  contains  two  decimal  pl&ces. 
Now  the  denominator  4— 2X.2 ;  and  since  2  is  contained  only  once  as  a  factor 


214  REDUCTION    OF  [SECT.   IX. 

in  10.  it  is  evident  that  10  must  be  repeated  as  many  times  as  a  factor  in  the 
numerator,  as  2  is  taken  times  as  a  factor  in  the  denominator,  in  order  to  re- 
duce the  fraction. 

For  the  same  reason  J  will  terminate  with  three  places,  and  is  equal  to  .125; 
for,  8.=2X2X2.  So  ^-—.2 ;  that  is,  the  decimal  terminates  with  one  place ;  for, 
since  its  denominator  5,  is  taken  only  mice  as  a  factor  in  10,  it  js  necessary  to 
add  only  one  cipher  to  its  numerator  in  order  to  reduce  it.  In  like  manner  it 
may  be  shown  that  the  number  of  figures  contained  in  any  terminate  decimal, 
w  equal  to  the  greatest  number  of  times  that  either  of  the  prime  factors  2  or  5, 
i*  repeated  in  the  denominator  of  the  given  fraction. 

The  same  reasoning  will  evidently  hold  true  when  the  numerator  is  2,  3,  4, 
&,  &c.,  or  any  number  greater  than  1.  In  this  case  the  decimal  will  be  as 
many  times  greater,  as  the  numerator  is  greater  than  1. 

344.  The  number  of  figures  in  the  period  must  always  be  owe 
less  than  there  are  units  in  the  denominator ;  for,  the  number  of 
remainders  different  from  each  other  which  can  arise  from  any 
operation  in  division,  must  necessarily  be  one  less  than  the  units 
in  the  divisor.  For  example,  in  dividing  by  7,  it  is  evident,  the 
only  possible  remainders  are  1,  2,  3,  4,  5,  and  6 ;  and  since  in  re- 
ducing a  common  fraction  to  a  decimal,  a  cipher  is  annexed  to 
each  remainder,  there  cannot  be  more  than  six  different  dividends  ; 
consequently,  there  cannot  be  more  than  six  different  figures  in 
the  quotient.  Thus,  -$-=.142857,142857,  &c- 

When  the  decimal  is  periodical  or  circulating,  it  is  custom- 
ary to  write  the  period  but  once,  and  put  a  dot,  or  accent  over 
the  first  and  last  figure  of  the  period  to  denote  its  continuance. 
Thus,  .46135135135,  &c.,  is  written  .46135,  and  .633333  &c.,  .63. 

Reduce  the  following  fractions  to  circulating  decimals : 

31.  i.  36.  f.  41.  f.  46,  £. 

32.  |.  37.  |.  42,  f.  47.  f. 

33.  i.  38.  f.  43.  i.  48.  £. 

34.  f.  39.  f.  44.  f.  49.  £. 

35.  'f.  40.  -f.  45.  f.  50.  -f. 

51.  How  many  decimal  figures  are  required  to  express  -^7 

52  How  many  decimal  figures  are  required  to  express  -fa  ? 

53  How  many  decimal  figures  are  required  to  express  ^$-5  ? 
54.  How  many  decimal  figures  are  necessary  to  express  yj-j  • 


ARTS.  344-346.]  DECIMALS.  215 

55    How  many  decimal  figures  are  necessary  to  express  -J-J-  ? 
66.  How  many  decimal  figures  are  necessary  to  express  ToW  • 

57.  Reduce  -&  to  a  decimal.        59.  Reduce  ^ir  to  a  decimal. 

58.  Reduce  -fe  to  a  decimal.        60.  Reduce  f£  to  a  decimal. 

Note.  —  For  the  method  of  finding  the  value  of  periodical  decimals,  or  of  re- 
ducing them  to  Common  Fractions,  also  of  adding,  subtracting,  multiplying  t 
and  dividing  them,  see  the  next  Section. 

CASE    III. 

345*   Compound  Numbers  reduced  to  Decimals. 
Ex.  1.  Reduce  13s.  6d.  to  the  decimal  of  a  pound. 

Analysis.  —  13s.  6d.=162d.,  and  £l  =  240d.  (Art.  281.)  Now 
162d.  is  i^  of  240d.  The  question  therefore  resolves  itself  into 
this  :  reduce  the  fraction  -J^  to  decimals.  Ans.  £.6*75.  Hence, 


340.  To  reduce  a  compound  number  to  the  decimal  of  a 
higher  denomination. 

First  reduce  the  given  compound  number  to  a  common  fraction  / 
(Art.  295  ;)  then  reduce  the  common  fraction  to  a  decimal.  (Art.  337.) 

2.  Reduce  4s.  9d.  to  the  decimal  of  £1.          Ans.  £.237+. 

3.  Reduce  10s.  9d.  to  the  decimal  of  £l. 

4.  Reduce  16s.  6d.  to  the  decimal  of  £l. 

5.  Reduce  17s.  7d.  to  the  decimal  of  £l. 

6.  Reduce  5d.  to  the  decimal  of  a  shilling. 

7.  Reduce  6£d.  to  the  decimal  of  a  shilling. 

8.  Reduce  37  rods  to  the  decimal  of  a  mile. 

9.  Reduce  ?  fur.  2  "ods  to  the  decimal  of  a  mile. 

10.  Reduce  j.o  ittm.  oO  «>ec.  to  tne  decimal  of  an  hour. 
1  1  .  Reduce  3  hrs.  3  min.  to  the  decimal  of  a  day. 

12.  Reduce  5  Ibs.  4  oz.  to  the  decimal  of  a  cwt. 

13.  Reduce  7  oz.  8  drams  to  the  decimal  of  a  pound. 

14.  Reduce  3  pks.  4  qts.  to  the  decimal  of  a  bushel. 

15.  Reduce  4  qts.  1  pt.  to  the  decimal  of  a  peck. 

16.  Reduce  4  qts.  1  pt  to  the  decimal  of  a  gallon. 

QUEST.—  346   How  is  a  compound  number  reduced  to  the  decimal  of  a  b\gher  dcnora 
inatiim  1 


216  REDUCTION    OF    DECIMALS.  [SECT.   IX. 

CASE    IV. 
347*  Decimal  Compound  Numbers  reduced  to  whole  ones. 

I.  Reduce  £.387  to  shillings,  pence  and  farthings. 

Operation.  Multiply  the  given  decimal  by  20,  because 
£.387  20s.  make  £l,  and  point  off  as  many  figures 
20  for  decimals,  as  there  are  decimal  places  in 
shil.  7.740  the  multiplier  and  multiplicand.  (Art.  330.) 
12  The  product  is  in  shillings  and  a  decimal 
pence  8.880  of  a  shilling.     Then  multiply  the  decimal 
4  of  a  shilling  by  12,  and  point  off  as  be- 
far.  3.520  fore,  &c.     The  numbers  on  the  left  of  the 
Ans.  7s.  8d.  3  far.  decimal  points,  viz :  7s.  8d.  3  far.,  form  the 
answer.     Hence, 

348*  To  reduce  a  decimal  compound  number  to  whole  num- 
bers of  lower  denominations. 

Multiply  the  given  decimal  by  that  number  which  it  takes  of  the 
next  lower  denomination  to  make  ONE  of  this  higher,  as  in  reduction, 
and  point  off  the  product,  as  in  multiplication  of  decimal  fractions. 
(Art.  330.)  Proceed  in  this  manner  with  the  decimal  figures  of 
each  succeeding  product,  and  the  numbers  on  the  left  of  the  decimal 
point  of  the  several  products,  will  be  the  whole  number  required. 

2.  Reduce  £.725  to  shillings,  pence  and  farthings. 

$,  Reduce  £.1325  to  shillings,  &c. 

4  Reduce  .125s.  to  pence  and  farthings. 

5.  Reduce  .825s.  to  pence  and  farthings. 

6.  Reduce  .125  cwt.  to  pounds,  &c, 

7.  Reduce  .435  Ibs.  to  ounces  and  drains. 

8.  Reduce  .275  miles  to  rods,  <fec. 

9.  Reduce  .4245  rods  to  feet,  &c, 

10.  Reduce  .1824  hhds.to  gallons,  &c. 

II.  Reduce  .4826  gal.  to  qts.,  &c. 

12.  Reduce  .4258  day  to  hours,  &c. 

13.  Reduce  .845  hr.  to  minutes  and  seconds. 

QnsT. — 3-18.  How  are  decimal  compound  numbers  reduced  to  whole  ones  of  a  lowar 
denomination! 


ARTS.  347-353.]     CIRCULATING  DECIMALS.  217 

SECTION    X. 
PERIODICAL    OR   CIRCULATING   DECIMALS.* 

ART.  349.  Decimals  which  consist  of  the  same  figures  or  set 
of  figures  repeated,  are  called  PERIODICAL,  OR  CIRCULATING  DECI- 
IIJLS.  (Art.  ,339.) 

35O.  The  repeating  figures  are  called  periods,  or  repetends. 
If  one  figure  only  repeats,  it  is  called  a  single  period,  or  repetend  ; 
as  .11111,  &c. ;  .33333,  &c. 

When  the  same  set  of  figures  recurs  at  equal  intervals,  it  is 
called  a  compound  period,  or  repetend;  as  .01010101,  &c. ; 
.123123123,  &c. 

351*  If  other  figures  arise  before  the  period  commences,  the 
decimal  is  said  to  be  a  mixed  periodical ;  all  others  are  called 
pure,  or  simple  periodicals.  Thus  .42C31G31,  &c.,  is  a  mixed 
periodical ;  and  .33333,  &c.,  is  a  pure  periodical  decimal. 

OBS.  1.  When  a  pure  circulating  decimal  contains  as  many  figures  as  there 
are  units  in  the  denominator  less  one,  it  is  sometimes  called  a  perfect  period, 
or  repetend,  (Art.  344.)  Thus,  i=.14'2857,  &c.,  and  is  a  perfect  period. 

2.  The  deciiral  figures  which  precede  the  period,  are  often  called  the  ter- 
minate part  of  the  fraction. 

352.  C;rculating  decimals  are  expressed  by  writing  the  period 
once  with  a  dot  over  its  first  and  last  figure  when  compound ;  and 
when  singl*  by  writing  the  repeating  figure  only  once  with  a  dot 
over  it.  Thus  .40135135,  &c.,  is  written  .40135  and  .33,  etc.,  .3. 

353^.  Similar  periods  are  such  as  begin  at  the  same  place  or 
distance  from  the  decimal  point ;  as  .1  and  .3,  or  2.34  and  3.70,  &c. 

•Dissimilar  periods  are  such  as  be^in  at  different  places ;  as 
.123  and  .42325 

Similar  and  conterminous  periods  are  such  as  begin  and  end  in 
the  same  places;  as  .2321  and  1034. 

*  Should  Periodical  Decimals  be  deemed  too  intricate  for  younger  classes,  they  can  be 
omitted  till  review 


218  PERIODICAL    OR  [SECT.   X 

REDUCTION   OP  CIRCULATING  DECIMA1  S. 

CASE  I. —  To  reduce  pure  circulating  decimals  to  common  frac- 
tions. 

354*  Tc  investigate  this  problem,  let  us  recur  to  the  origin 
of  circulating  decimals,  or  the  manner  of  obtaining  them.  For 
example,  -£=.11111,  <fec.,  or  .1 ;  therefore  the  true  value  of  .11111, 
&c.,  or  .1,  must  be  i  from  which  it  arose.  For  the  same  reason 
.22222,  &c.,  or  .2,  must  be  twice  as  much  or  f ;  (Art.  186;) 
.33333,  (fee.,  or  .3=f;  .4=f;  .5=%,  (fee. 

Again,  -gV=.010101,  &c.,  or  .61  ;  consequently  .010101,  &c.,  01 
.oi^-g-V;    .020202,  <fec.,  or  .02=-&;  .030303,  <fec.,  or  .03=tV 
.070707,  &c.,  or  .67=VV>  &c.     So  also  -9-^=. 00 100 1001,  &c. 
or  .6oi  ;  therefore  .001001,  &c.,  or  .6oi=-9-h- ;  .602=-9-f-9  ;  &c. 

In  like  manner  £=.142857  ;  (Art.  337;)  and  i42857=iiltf|; 
for,  multiplying  the  numerator  and  denominator  of  £  by  142857, 
we  have  iff  Iff .  (Art.  191.)  So  f  is  twice  as  much  as  £ ;  f,  three 
times  as  much,  &c.  Thus  it  will  be  seen  that  the  value  of  a  pure 
periodical  decimal  is  expressed  by  the  common  fraction  whose 
numerator  is  the  given  period,  and  whose  denominator  is  as  many 
9s  as  there  are  figures  in  the  period.  Hence, 

355.  To  reduce  a  pure  circulating  decimal  to  a  common 
fraction. 

Make  the  given  period  the  numerator,  and  the  denominator  will 
be  as  many  9s  as  there  are  figures  in  the  period. 

Ex.  1.  Reduce  .3  to  a  common  fraction.  Ans.  f,  or  f. 

2.  Reduce  .6  to  a  common  fraction.  Ans.  -f,  or  |. 

3.  Reduce  .18  to  a  common  fraction. 

4.  Reduce  .123  to  a  common  fraction. 

5.  Reduce  .297  to  a  common  fraction. 

6.  Reduce  .72  to  a  common  fraction. 

7.  Reduce  .09  to  a  common  fraction. 

8.  Reduce  .045  to  a  common  fraction. 

9.  Reduce  .142857  to  a  common  fraction. 
10.  Reduce  .076923toa  common  fraction 


ARTS.  354—  357.  J     CIRCULATING  DECIMALS.  219 

CASE  II.  —  To  i  educe  mixed  circulating  decimals  to  common 
fractions. 

3  50.  11.  Reduce  .16  to  a  common  fraction. 

Analysis.  —  Separating  the  mixed  decimal  into  its  terminate  and 
periodical  part,  we  have  .16  =  .!  +.06.  (Art.  320.)  Now  .1  =fa  ; 
(Art.  312  ;)  and  .06=-&;  for,  the  pure  period  .6=f,  (Art.  351,) 
and  since  the  mixed  period  .06,  begins  in  hundred  ths*  place,  its 
value  is  evidently  only  fa  as  much  ;  but  f-MO=VV-  (Art.  227.) 
Therefore  .16=-tV+"9Io-  Now  -fa  and  -9\,  reduced  to  a  common 
denominator  and  added  together,  make  •£•$,  or  £.  Ans. 

OBS.  In  mixed  circulating  decimals,  if  the  period  begins  in  hundredths'  place 
it  is  evident  from  the  preceding  analysis  that  the  value  of  the  periodical  part  is 
only  -fa  as  much  as  it  would  be,  if  the  period  were  pure  or  begun  in  tenths' 
place  ;  when  the  period  begins  in  thousandths'  place,  its  value  is  only  -j-J-g-  part 
as  much,  &c.  Thus  .6=f  ;  .06=f-s-10=-^-0  :  .00(i=f-j-lOO=rfir,  &c. 

357.  Hence,  the  denominator  of  the  periodical  part  of  a 
mixed  circulating  decimal,  is  always  as  many  Os  as  there  are  fig- 
ures in  the  period  with  as  many  cipher*  annexed  as  there  are  deci- 
mals in  the  terminate  part. 

12.  Reduce  .8567923  to  a  commop  fraction. 

Solution.  —  Reasoning  as  before  .8567923=-^ftr+9?n»ufl'  Re- 
ducing these  two  fractions  to  the  least  common  denominator, 
(Art.  261.)  t\*oX99909H^-ilHru  whose  denominator  is  the  same 
as  that  of  the  other.  Now 


Contraction.  To  multiply  by  99999,  annex  as  many 

8500000  ciphers  to  the  multiplicand  as  there  are 

85  9s  in  the  multiplier,  &c.  (Art.  105.)    This 

8499915  1st  Nu.       gives  the  numerator  of  the  first   fraction 

67923  2d  Nu.        or  terminate  part,  to  which  add   the  nu- 

8567838  merator  of  the  second  or  periodical  part, 

9999900  Ans.  and  the  sum  will  be  the  numerator  of  the 

answer.     The  denominator  is  the  same  as 

that  of  the  second  or  periodical  part. 

10* 


220  PERIODICAL    OR  [SECT.  X. 

Second  Method. 

8567923  the  given  circulating  decimal. 

85  the  terminate  part  which  is  subtracted 
8567838  the  numerator  of  the  answer. 


Note.  —  1.  The  reason  of  this  operation  may  be  shown  thus:  8567923ra 
8500000-J-G7923  Now  8500000—  85+67923  is  equal  to  8567923-  -85. 

2.  Jt  is  evident  that  the  required  denominator  is  the  same  as  that  of  the 
periodical  part;  (Art.  357;)  for,  the  denominator  of  the  periodical  part  is  the 
least  common  multiple  of  the  two  denominators.  Hence, 

358.  To  reduce  a  mixed  circulating  decimal  to  a  common 
fraction. 

Change  Loth  tlie  terminate  and  periodical  part  to  common  frac- 
tions 'separately,  and  their  sum  will  be  the  answer  required. 

Or,  from  the  given  mixed  periodical,  subtract  the  terminate  part, 
and  tlie  remainder  will  be  the  numerator  required.  The  denominator 
is  always  as  many  9s  as  there  are  figures  in  the  period  with  as 
many  ciphers  annexed  as  there  are  decimals  in  the  terminate  part. 

PROOF.  —  Change  the  common  fraction  back  to  a  decimal,  and  if  the 
result  is  the  same  as  the  given  circulating  decimal,  the  work  is  right. 

13.  Reduce  .138  to  a  common  fraction.     Ans.  -J-JHJ-,  or  -^-. 

14.  Reduce  .53  to  a  common  fraction. 

15.  Reduce  .5925  to  a  common  fraction. 

16.  Reduce  .583  to  a  common  fraction. 

17.  Reduce  .0227  to  a  commmon  fraction. 

18.  Reduce  .4745  to  a  common  fraction. 

19.  Reduce  .5925  to  a  common  fraction. 

20.  Reduce  .008497133  to  a  common  fraction. 

CASE  III.  —  Dissimilar  periodicals  reduced  to  similar  and  confer* 
minous  ones. 

359*  In  changing  dissimilar  periods,  or  repetends,  to  similar 
and  conterminous  ones,  the  following  particulars  require  attention. 

1,  Any  terminate  decimal  may  be  considered  as  interminate 
by  annexing  ciphers  continually  to  the  numeratoj  .     Thus  .46= 
.460000,  &c.=.460 


ARTS.  358-361. J     CIRCULATING  DECIMALS.  221 

2.  Any  pure  periodical  may  be  considered  as  mixed,  by  taking 
the  given  period  for  the  terminate  part,  and  making  the  given 
period  the  intcrminate  part.     Thus  .4G=.4G  +  .004G,  (fee. 

3.  A  single  period  may  be  regarded  as  a  compound  periodical. 
Thus  .3  may  become  .33,  or  .333 ;  so  .63  may  be  made  .0333, 
or  .03033,  &c. 

4.  A  single  period  may  also  be  made  to  begin  at  a  lower  order, 
regarding  its  higher  orders  as  terminate  decimals.     Thus  .3  may 
be  made  .33,  or  .3333,  (fee. 

5.  Compound  periods  may  also  be  made  to  begin  at  a  lower 
order.     Thus  .30  may  be  changed  to  .303,  or  .30303,  (fee. ;  or  by 
extending  the  number  of  .places  .479  may  be  made  .47979,  or 
.4797979,  <fec. ;  or  making  both  changes  at  once  .532  may  be 
changed  to  .5325325,  (fee.     Hence, 

3 GO.  To  make  any  number  of  dissimilar  periodical  decimals 
similar. 

Move  the  points,  so  that  each  period  shall  begin  at  the  same  order 
ar>  the  period  which  has  the  most  figures  in  its  terminate  part. 

21.  Change  0.814,  3.20,  and  .083  to  similar  and  conterminous 
periods. 

Operation.  Having  made  the  given  periods  simi- 

0. 814  =  0. 81481481         lar,  the  next  step  is  to  make  them  con- 

3.20  =  3.20202026         terminous.     Now  as  one  of  the  given 

.083  =  0.08333333         periods  contains  3  figures,  another  2, 

and  the  other  1,  it  is  evident  the  new 

periodical  must  contain  a  number  of  figures  which  is  some  multi- 
ple of  the  number  of  figures  in  the  different  periods;  viz:  3,  2, 
and  1.  But  the  least  common  multiple  of  3,  2,  and  1  is  6  ;  there- 
fore the  new  periods  must  at  least  contain  0  figures.  Hence, 

36  !•  To  make  any  number  of  dissimilar  periodical  decimals, 
similar  and  conterminous. 

Firm  make  the  periods  similar  ;  (Art.  300;)  then  extend  the  fig- 
ure;>  of  each  to  as  many  places,  as  there  are  units  in  the  le^st  com- 
mon multiple  of  the  NUMBER  of  periodical  fiyures  contained  in  each 
\>f  the  given  decimals.  (Art.  176.) 


222  PERIODICAL    OR  T^ECT.  X. 


L' 

22.  Change  46.162,  5.26,  63.423,  .486,  and  12.5,  to  similar 
and  conterminous  periodicals. 

Operation. 

46.162  =  46.16216216  The  numbers  of  periodical  figures  in 

5.26   =   5.26262626  the  given  decimals  are  3,  2,  3,  and  1 ; 

03.423  =  63.42342342  and    the   least    common   multiple   of 

.486=   0.48666666  them  is  6.     Therefore  the  new  periods 

125     =12.50600006  must  each  have  6  figures; 

23  Make  .27,  .3,  and  .045  similar  and  conterminous. 

24  Make  4.321,  6.4263,  and  .6  similar  and  conterminous. 

ADDITION  OF  CIRCULATING  DECIMALS. 

Ex.  1.    What   is    the   sum   of   17. 23 +41. 2476  +  8. 61+1.5  + 
35.423  ? 

Operation. 

Dissimilar.      Sim.  &.  Conterminous. 

17.23     =17.2323232  First  make  the  given  decimals  sim- 

41.2476  =  41.2476476       ilar  and  conterminous.     (Art.  361.) 

%8.61     =   8.6161616       Then  add  the  periodical  parts  as  in 

1.5        =   1.5000006       simple  addition,  and  since  there  are 

35.423   =35.4232323       6  figures  in  the  period,  divide  their 

Ans.  104.0i93648       sum  D7  999999;  for  this  would  be 

its  denominator,  if  the  sum  of  the 

periodicals  were  expressed  by  a  common  fraction.  (Art.  355.) 
Setting  down  the  remainder  for  the  repeating  decimals,  carry  the 
quotient  1  to  the  next  column,  and  proceed  as  in  addition  of 
whole  numbers..  Hence, 

362*  We  derive  the  following  general 

RULE   FOR  ADDING  CIRCULATING  DECIMALS. 

first  make  the  periods  similar  and  conterminous,  and  find  iheij 
^um  as  in  Simple  Addition.  Divide  this  sum  by  as  me  ny  9s  as 
there  are  figures  in  the  period,  set  the  remainder  under  the  figures 
added  for  the  period  of  the  sum,  carry  the  quotient  to  the 
column,  and  proceed  with  the  rest  as  in  Simple  Addition. 


ARTS.  362,  363.]     CIRCULATING  DECIMALS.  223 

OBS.  If  the  remainder  has  not  so  many  figures  as  the  period,  ciphers  must 
be  prefixed  to  make  up  the  deficiency. 

2.  What  is  the  sum  of  24.132  +  2. 23+85.24  +  67.6  ? 

3.  What  is  the  sum  of  328.126+81.23+5.624+61.6*? 

4.  What  is  the  sum  of  31.62  +  7.824  +  8.392+.027  ? 

5.  What  is  the  sum  of  462.34  +  60.82  +  71.164  +  .35  ? 

6.  What  is  the  sum  of  60.25+.84+6.435+.45+45.24  ? 

7.  What  is  the  sum  of  9.814+1.5+87.26+0,83+124.09? 

8.  What  is  the  sum  of  3. 6  +  78.3476  +  735. 3+.37S  +  . 27  + 
187.4? 

9.  What  is  the  sum  of  5391.357+72.38+187.21+4.2965  + 
217.8496+42.176+.523+58.30048? 

10.  What  is  the  sum  of  .  162  +  134.09+2. 93  +  97. 26  +  3.769°-SO 
+99.083+1.5+.814? 

SUBTRACTION  OP   CIRCULATING  DECIMALS. 
Ex.  1.  From  52.86  take  8.37235. 

Operation.  We  first  make  the  given  decimals  sn  i- 

52.86=52.86868       lar  and  conterminous,  then  subtract  as   n 

8.37235=  8.37235       whole  numbers.     But  since  the  period  tn 

44.49632       tne  lower  line  is  larger  than  that  above 

it,  we  must  borrow  1  from  the  next  higher 

order.     This  will  make  the  right  hand  figure  of  the  remainder  one 
less  than  if  it  was  a  terminate  decimal.     Hence, 

363*  We  derive  the  following  general 

RULE   FOR   SUBTRACTING  CIRCULATING   DECIMALS. 

Make  the  periods  similar  and  conterminous,  and  subtrttt  as  in, 
ivhole  numbers.  If  the  period  in  the  lower  line  is  larger  than  that 
above  it,  diminish  the  right  hand  figure  of  the  remainder  ^y  1. 

OBS.  The  reason  for  diminishing  the  right  hand  figure  of  the  remainder  by 
1,  if  -he  period  in  the  lower  line  is  larger  than  that  above  it,  may  be  explained 
thus: 

When  the  period  in  the  lower  l;ne  is  larger  than  that  above  it,  we  must  evi- 
dently borrow  1  from  the  next  higher  order.  Now  if  the  given  decimals  were 
extended  to  a  second  period,  in  this  period  the  lower  number  would  also  bo 


224  PERIODICAL    OR  [SECT.   X. 

larger  than  that  above  it,  and  therefore  we  must  borrow  1.  But  having  bor- 
rowed 1  in  the  seoDnd  period,  we  must  also  carry  one  to  the  next  figure  in  the 
lower  line,  or,  what  is  the  same  in  effect,  diminish  the  right  hand  figure  of  the 
remainder  by  1. 

2.  From  85.62  take  13.76432.     Ans.  71.86193. 

3.  From  476.32  take  84.7697. 

4.  From  3.8564  take  .0382. 

5.  From  46.123  take  41.3. 

6.  From  801.6  take  400.75. 

7.  From  4.7824  take  .87. 

8.  From  1419.6  take  1200.9. 

9.  From  .634852  take  .021. 

10.  From  8482.421  take  6031.035. 


MULTIPLICATION  OF   CIRCULATING  DECIMALS 
Ex.  1.  What  is  the  product  of  .36  into  .25  ? 

Operation.  We  first  reduce  the  given  periodi- 

.36=ff=-^f  cals  to  common  fractions  ;  (Art.  357  ;) 

.25  =  -[\+95u=-ff.  then   multiply  them   together.    (Art 

Now  T4rXli=-999^  219.)     Finally,  we  reduce  the  product 

But  ^j^=.092  Ans.  to  a  periodical  decimal.     Hence, 

364*  We  derive  the  following  general 

RULE  FOR  MULTIPLYING  CIRCULATING  DECIMALS. 

First  reduce  the  given  periodicals  to  common  fractions,  and  mul- 
tiply them  together  as  usual.  (Art.  219.)  Finally,  reduce  the  prod- 
uct to  decimals  and  it  will  be  the  answer  required. 

OBS.  If  the  numerators  and  denominators  have  common  factors,  the  opera- 
tion may  be  contracted  by  canceling  those  factors  before  the  multiplication  ifl 
psrformet:.  (Art.  221.) 

2.  What  is  the  product  of  37.23  into  .26  ?     Ans.  9.928. 

3.  What  is  the  product  of  .123  into   6  ? 

4.  What  is  the  product  of  .245  into  7.3  ? 

5.  What  is  the  product  of  24.6  into  15.7  ? 

6.  What  is  the  product  of  48.23  into  16.13  ? 


ARTS.  364.  365.J     CIRCULATING  DECIMALS. 

7.  What  is  the  product  of  8574.3  into  87.5  ? 

8.  What  is  the  product  of  3.973  into  .8  ? 

9.  What  is  the  product  of  49640.54  into  .70503  ? 
10.  What  is  the  product  of  7.72  into  .297  ? 

DIVISION  OF   CIRCULATING  DECIMALS. 

Ex.  1.  Divide  234.6  by  .7. 

Operation.  We  first  reduce   the   divisor 

234.6  =  234-f=-2L|A  and  dividend  to  common  frac- 

.7=|-  tions;    (Art.  357 ;)    and    divide 

Now  -H^i^-H^Xf^-ft3-       one  by  the  other;    (Art.  229  ;) 

And  -^^=301.714285  Ans.      then  reduce  the  quotient  to  a 

decimal.   (Art.  337.)     Hence, 

365*  We  derive  the  following  general 

RULE   FOR  DIVIDING  CIRCULATING  DECIMALS. 

Reduce  the  divisor  and  dividend  to  common  fractions ;  divide 
one  fraction  by  the  other,  and  reduce  the  quotient  to  decimals. 

OBS.  After  the  divisor  is  inverted,  if  the  numerators  and  denominator*  have 
factors  common  to  both,  the  operation  may  be  contracted  by  canceling  those         , 
fectors.  (Art.  232.) 

2.  Divide  319.28007112  by  764.5.     Ans.  0.4176325. 

3.  Divide  18.56  by  .3. 

4.  Divide  .6  by  .123. 

5.  Divide  2.297  by  .297. 

6.  Divide  750730.518  by  87.5. 

7.  Divide  42630.6  by  28421.3. 

8.  Divide  80000.27  by  20000.36. 

9.  Divide  24.081  by  .386. 
10.  Divide  .36  by  .25. 


226  REDUCTION    OP  fSfiCT.  X] 

SECTION    XI. 
FEDERAL    MONEY. 

ART.  366*  FEDERAL  MONEY  is  the  currency  of  the  United 
States.  Its  denominations,  we  have  seen,  are  Eayles,  Dollars, 
.Dimes,  Cents,  and  Mills.  (Art.  244.) 

36 7 •  All  accounts  in  the  United  States  are  required  by  law 
to  be  kept  in  dollars,  cents,  and  mills.  Eagles  are  expressed  in 
dollars,  and  dimes  in  cents.  Thus,  instead  of  8  eagles,  we  say, 
80  dollars ;  instead  of  6  dimes  and  7  cents,  we  say,  67  cents,  &c. 

368.  Federal  Money  is  based  upon  the  Decimal  system  of 
Notation.  Its  denominations  increase  and  decrease  from  right  to 
left  and  left  to  right  in  a  tenfold  ratio,  like  whole  numbers  and 
decimals.  (Art.  244.  Obs.  1.) 

369*  The  dollar  is  regarded  as  the  unit ;  cents  and  mills  are 
fractional  parts  of  the  dollar,  and  are  distinguished  from  it  by 
a  decimal  point  or  separatrix  (.)  in  the  same  manner  as  common 
decimals  are  distinguished  from  whole  numbers.  (Art.  311.) 
Dollars  therefore  occupy  units'  place  of  simple  numbers ;  eayles, 
or  tens  of  dollars,  tens'  place,  &c.  Dimes,  or  tenths  of  a  dollar, 
occupy  the  place  of  tentlis  in  decimals  ;  cents,  or  hundred ths  of  a 
dollar,  the  place  of  hundredths  ;  mills,  or  thousandths  of  a  dollar, 
the  place  of  thousandths  ;  tenths  of  a  mill,  or  ten  thousandths  of 
a  dollar,  the  place  of  ten  thousandths,  &c. 

OBS.  1.  Since  dimes  in  business  transactions  are  expressed  in  cents,  two  places 
of  decimals  are  assigned  to  cents.  If  therefore  the  number  of  cents  is  less  than 
10,  a  cipher  must,  always  be  placed  on  the  left  hand  of  them ;  for  cents  are  hun- 
dredths of  a  dollar,  and  hundredths  occupy  the  second  decimal  place.  (Art.  313.) 
For  example,  4  cents  are  written  thus  .04 ;  7  cents  thus  .07;  &c. 

2.  Mills  occupy  the  third  place  of  decimals ;  for  they  are  thousandths  of  e 
dollar.  Consequently,  when  there  are  no  cents  in  the  given  sum,  two  cipher 
must  be  placed  before  the  mills.  Hence, 

QUEST.— 366.  What  is  Federal  Money?    367.  In  what  are  accounts  kept  in  the  U.  S. 
How  would  you  express  8  eagles  ?     How  express  6  dimes  and  "  cents  ?    3f>8.  Upon  what  i.. 
Federal  Money  based  1    369.  What  is  regarded  as  the  unit  ir  Federal  Money  ?    What  aiv 
jents  and  mills  ?    How  are  they  distinguished  from  dollars  ? 


ARTS.  386-371. J          FEDERAL  MONEY.  227 

37O.  To  read  any  sum  of  Federal  Money. 

Call  all  the  figures  on  the  left  of  the  decimal  point  dollars  ;  the 
first  two  figures  after  the  point,  are  cents  ;  the  third  figure  denotes 
mills  ;  the  other  places  on  the  right  are  decimals  of  a  mill.  Thus, 
$3.25232  is  read,  3  dollars,  25  cents,  2  mills,  and  32  hundredth^ 
of  a  mill. 

OBS.  Sometimes  all  the  figures  after  the  point  are  read  as  decimals  of  a  :ol- 
ar,  Thus,  $5.356  is  read,  "  5  and  356  thousandths  dollars." 

Write  the  following  sums  in  Federal  money : 

1.  70  dollars,  and  8  cents.  Ans.  $70.08. 

2.  150  dollars,  3  cents,  and  5  mills. 

3.  409  dollars,  40  cents,  and  3  mills. 

4.  200  dollars,  5  cents,  and  2  mills. 

5.  4050  dollars,  65  cents,  and  3  mills. 

Note. — In  business  transactions,  when  dollars  and  cents  are  expressed  to- 
gether, the  cents  are  frequently  written  in  the  form  of  a  common  fraction. 
Thus,  the  sum  of  $75.45,  is  written  75-^^  dollars. 


REDUCTION  OP  FEDERAL  MONEY. 

CASE    I. 
Ex.  1.  How  many  cents  are  there  in  95  dollars? 

Solution. — Since  in  1  dollar  there  are  100  cents,  in  95  dollars 
there  are  95  times  as  many.     And  95  X  100=9500. 

Ans.  9500  cents. 
2.  In  20  cents  how  many  mills  ?  Ans.  200  mills. 

Note. — To  multiply  by  10,  100,  &c.,  we  simply  annex  as  many  ciphers  to 
Ihe  multiplicand,  as  there  are  ciphers  in  the  multiplier.  (Art.  99.)  Hence, 

3  7  1  •   To  reduce  dollars  to  cents,  annex  two  ciphers. 
To  reduce  dollars  to  mills,  annex  three  ciphers. 
To  reduce  cents  to  mills,  annex  one  cipher. 

OBS.  To  reduce  dollars,  cents,  and  mills,  to  mills,  erase  the  sign  of  dollars 
and  the  scparatrix.  Thus,  $25,36  reduced  to  cents,  becomes  2536  cents. 

QUEST.— 353.  How  do  you  read  Federal  Money  1  Obs.  What  other  mode  of  reading 
Federal  Money  is  mentioned  1  354.  How  are  dollars  reduced  to  cents  ?  Dollars  to  mills? 
Cents  to  mills?  Obs.  Dollars,  cents,  and  mills,  to  mills? 


228  ADDITION  op  [SECT.  XI 

3.  In  $12  l:ow  man)-  cents?  Ans.  1200  cents. 

4.  In  $460  how -many  cents? 

5.  In  $95  how  many  mills  ? 

6.  In  90  cents  how  many  mills  ? 

7.  Reduce  $25.15  to  cents. 

8.  Reduce  $864.08  to  cents. 

9.  Reduce  $1265.05  to  mills. 

10.  Reduce  $4580.10  to  mills. 

11.  Reduce  $6886.258  to  mills. 

12.  Reduce  $85625.40  to  mills. 

CASE    II. 

13.  In  6400  cents,  how  many  dollars? 

Suggestion. — Since  100  cents  make  1  dollar,  6400  cents  will 
make  as  many  dollars  as  100  is  contained  times  in  6400.  And 
6400-MOO  =  64.  Am.  $64. 

14.  In  260  mills,  how  many  cents?  Ans.  26  cents. 

Note. — To  divide  by  10, 100,  &c.,  we  simply  cut  off  as  many  figures  from  the 
right  of  the  dividend  as  there  are  ciphers  in  the  divisor.  (Art.  131.)  Hence, 

372.  To  reduce  cents  to  dollars,  cut  off  two  figures  on  the 
right. 

To  reduce  mills  to  dollars,  cut  off  three  figures  on  the  riyht. 
To  reduce  mills  to  cents,  cut  off  one  figure  on  the  right. 
OBS.  The  figures  cut  off  are  cents  and  mills. 

15.  In  626  cents,  how  many  dollars?       Ans.  $6.26. 

16.  In  1516  cents,  how  many  dollars? 

17.  In  162  mills,  how  many  cents? 

18.  In  1000  mills,  how  many  dollars? 

19.  In  2360  mills,  how  many  cents  ? 

20.  In  3280  mills,  how  many  dollars? 
21  Reduce  8500  cents  to  dollars. 

22,  Reduce  2345  cents  to  dollars,  &c. 

23.  Reduce  92355  mills  to  dollars,  &c. 

QC«ST.— 355.  How  are  cents  reduced  to  dollars?  Mills  to  dollars!  Mills  to  cental 
Ob*.  What  are  the  figures  cut  off  1 


ARTS.  372-374.]          FEDERAL  MONEY.  229 

24.  Reduce  150233  mills  to  dollars,  &c. 

25.  Reduce  450341  cents  to  dollars,  &c. 

373.  Since  Federal  Money  is  expressed  according  to  the  de- 
cimal system  of  notation,  it  is  evident  that  it  may  be  subjected  to 
the  same  operations  and  treated  in  the  same  manner  as  Decimal 
fractions. 


ADDITION  OF  FEDERAL   MONEY. 

Ex.  1.  A  man  bought  a  cloak  for  $35.375,  a  hat  for  $4.875,  \ 
pair  of  boots  for  $6.50,  and  a  coat  for  $23.625  :  what  did  he  pay 
for  all? 

Operation.  We  write  the  dollars  under  dollars,  cents 

$35.375  under  cents,  &c.    Then  add  each  column  sepa- 

4.875  rately,  and  point  off  as  many  figures  for  cents 

6.50  and  mills,  in  the  amount,  as  there  are  places 

23.625  of  cents  and  mills  in  either  of  the  given  num~ 

$70.375  Am.  bers.     Hence, 

374.  We  derive  the  following  general 

RULE   FOR  ADDING   FEDERAL   MONEY. 

Write  dollars  under  dollars,  cents  under  cents,  &c.,  so  that  the 
same  orders  or  denominations  may  stand  under  each  other.  Add 
each  column  separately,  and  poini  off  tlie  amount  as  in  addition  of 
decimal  fractions.  (Art.  320.) 

OES.  If  either  of  the  given  numbers  have  no  cents  expressed,  it  is  customary 
to  supply  their  place  by  ciphers. 

2.  What  is  the  sum  of  $48.25,  $95.60,  $40.09,  and  $81.10  ? 

3.  What  is  the  sum  of  $103.40,  $68.253,  $89.455,  $140  02, 
and  $180? 

4.  What  is  the  sum   of  $136.255,   $10.30,  $248.50,  $65.33, 
and  $100.125? 

QUEST. — 357.  How  is  Federal  Money  added  1  How  point  off  the  amount  Obi.  When 
»ny  of  the  given  numbers  have  no  cents  expressed,  how  is  their  place  suppMed  ? 


230  SUBTRACTION    OP  [SECT.  XL 

5.  What  is  the  sum  of  $170,   $400.02,   8130,  $250.10,  ana 
$845.22? 

6.  What  is  the  sum  of  $268,45,  $800.05,  $192.125,-  $80.625, 
and  $90.25  ? 

7.  What  is  the  sura  of  $1500.20,  $1050.07,  $100.70,  $95.025, 
$360.437,  and  $425? 

8.  What  is  the  sum  of  $2600,  $1927.404,  $1603.40,  $3304.17 
$165.47,  and  $2600.08? 

9.  A  man  bought  a  load  of  hay  for  $19.675,  a  horse  for  $73  25, 
a  yoke  of  oxen  for  $69.56,  a  cow  for  $17,  and  a  calf  for  $5.80: 
what  did  he  pay  for  all  ? 

10.  A  lady  gave  $21.50  for  a  dress,  $9.25  for  a  bonnet,  $28.33 
for  a  shawl,  and  $15.25  for  a  muff:  what  was  her  bill? 

11.  A  jockey  bought  a  span  of  horses  for  $276.87,  and  sold 
them  so  as  to  gain  $73.45 :  how  much  did  he  sell  them  for? 

12.  A  man  gave  $4925.68  for  a  farm,  and  sold  it  so  as  to  gain 
$1565.37  :  how  much  did  he  sell  it  for? 

13.  A  man  sold  a  sloop  for  $7623.87,  which  was  $1141.25  less 
than  cost :  how  much  did  it  cost  ? 

14.  A  man  bought  a  block  of  stores  for  $15268,  which  was 
$1721  less  than  cost:  what  was  the  cost? 

15.  What  is  the  sum  of  134  dolls.  3  cts.  7  mills,  108  dolls. 
6  cts.  8  mills,  90  dolls.  9  cts.  4  mills,  and  46  dolls.  18  cts.  4  mills  ? 

'       16.  What  is  the  sum  of  61  dolls.  1  ct.  2  mills,  19  dolls.  11  cts. 
4  mills,  140  dolls,  and  80  dolls.  4  cts.? 

17.  What  is  the  sum  of  140  dolls.  10  cts.,  69  dolls.  3  cts. 
8  mills,  18  dolls.  7  cts.,  and  29  dolls.  5  mills? 

18.  What  is  the  sum  of  860  dolls.  8  cts.,  298  dolls.  4  cts.  8  mills, 
416  dolls.,  280  dolls.  13  cts.,  and  91  dolls.  ? 

19.  What  is  the  sum  of  14209  dolls.,  65241  dolls.,  1050  dolls., 
610  dolls.  7  cts.,  and  1000  dolls.  10  cts.  ? 

20.  What  is  the  sum  of  1625  dolls.,  4025  dolls.,  1863  dolls. 
75  cts.,  16000  dolls.,  and  48261  dolls.? 

21.  What  is  the  sum  of  8  thousand  dolls.,  2  hundred  and  60 
dolls.  5  cts.,  19  thousand  dolls.  60  cts.,  6  hundred  dolls.  9  cts.? 

22.  What  is  the  sum  of  19  thousand  dolls.  50  cts.,  61  thou- 
sand dolls.  10  cts.,  18  hundred  dolls.  3  ctf .  ? 


ART.  375.]  FEDERAL  MONEY. 

SUBTRACTION   OF    FEDERAL   MONEY. 

Ex.  1.  A  merchant  bought  a  quantity  of  molasses  for  $75.40; 
and  a  box  of  sugar  for  $42.63  :  how  much  more  did  he  pay  for 
one  Uian  the  other  ? 

Operation.  We  write  the  less  number  under  the  greater, 

$75.40  placing  dollars  under  dollars,  &c.,  then  subtract 

42.63  and  point  off  the  answer,  as  in  subtraction  of 

$32.77  decimals.     Hence, 

375*  We  derive  the  following  general 

RULE   FOR   SUBTRACTING   FEDERAL   MONEY. 

Write  the  less  number  under  the  greater,  with  dollars  under  dol- 
lars, cents  under  cents,  &c. ;  then  subtract  and  point  off  the  remain- 
der as  in  subtraction  of  decimal  fractions.  (Art.  322.) 

OBS  If  either  of  the  given  numbers  have  no  cents  expressed,  it  is  customary 
to  supply  their  place  by  ciphers. 

2.  A  man  bought  a  horse  for  175.50,  and  sold  it  for  $87.63* 
how  much  did  he  make  by  his  bargain? 

3.  If  a  man  deposits  $204.65  in  a  bank,  and  afterwards  checks 
out  $119.83,  how  much  will  he  have  left  ? 

4.  A  man  owing  $682.40,  paid  $435.25 :    how  much  does  ha 
still  owe  ? 

5'.  A  man  owing  $982.68,  paid  all  but  $64.20  :  how  much  did 
he  pay  ? 

6.  A  merchant  bought  a  quantity  of  goods  for  $833.63,  and  re- 
tailed them  for  $1016.85  :  how  much  did  he  make  by  the  bargain  ? 

7.  A  merchant  bought  a  lot  of  goods  for  $1265.82,  and  sold 
them  for  $942.35  :  how  much  did  he  lose  ? 

8.  A  grocer  sold  a  lot  of  sugar  for  $635.20,  and  made  thereby 
$261 ,38  :  how  much  did  he  pay  for  the  sugar  ? 

9.  A  man  sold  his  farm  for  $12250.62,  which  was  $1379.87 
more  than  it  cost :  how  much  did  it  cost  ? 


QUEST. — 358.    How  is  Federal  Money  subtracted  ?     How  point  off  the  remainder  1 
Obs    When  either  of  the  given  numbers  have  no  cents,  how  is  their  place  supplied  1 


232  MULTIPLICATION    OP  [SECT.   XJ 

10.  From  $10600.75  take  $8901.26. 

11.  From  $20206.85  take  $10261.062. 

12.  From  $61219.40  take  $100.036. 

13.  From  $19  take  1  cent  and  9  mills. 

14.  From  89  dollars  take  89  cents. 

15.  From  506  dolls,  take  316  dolls,  and  i  cts. 

1 6.  From  5  dolls.  7  mills  take  2  dolls.  7  els. 

17.  From  61  dolls.  6  cts.  take  29  dolls.  4  mills. 

18    From  11000  dolls.  10  cts.  take  110  dolls.  3  cts. 
19.  From  1 00 100  dolls,  take  10110  dolls.  10  cts. 


MULTIPLICATION   O^    FEDERAL   MONEY. 

376.  In  multiplication  of  Federal  Money,  as  well  as  in  simple 
numbers,  the  multiplier  must  always  be  considered  an  abstract 
number.  (Art.  82.  Obs.  2.) 

Ex.  1.  What  will  8  bbls.  of  flour  cost,  at  $5.62  per  bbl.  ? 

Analysis.  —  Since  1  bbl.  costs  $5.62,  8  bbls.  will  cost  8  times  as 
much;  and  $5.62  X8=$44.96  Am. 

2.  What  cost  21.7  bushels  of  apples,  at  15  cts.  per  bushel? 

Operation.  Reasoning  as  before,  21.7  bushels  will  cost 

21.7  21.7   times    15   cents.     But  in  performing  the 

.15  multiplication,  it  is   more  convenient  to  make 

1085  the  .15  the  multiplier,  and  the  result  will  be 

217  the  same  as  if  it  was  placed  for  the  multipli- 

$3.255  Ans.  cand.  (Art.  83.)     Point  off  the  product  as  be- 

fore.    Hence, 

377.  When  the  price  of  one  article,  one  pound,  one  yard,  &c.,  is 
given,  to  find  the  cost  of  any  number  of  articles,  pounds,  yards,  tfec. 

Multiply  the  price  of  one  article  and  the  number  of  articles  to- 
gether, and  point  off  the  product  as  in  multiplication  of  decimal*, 
(Art.  324.) 


In  Multiplication  of  Federal  Money,  what  must  one  of  the  given  factor* 
be  considered  1  377.  When  the  price  of  one  article,  one  pound,  &c.,  is  given,  how  is  the 
cost  of  any  number  of  articles  found  1 


ARTS.  376-378.]         FEDERAL  MONLY.  233 

3.  What  cost  17.6  yards  of  cloth,  at  $4.75  per  yard? 

4.  Multiply  $25.625  by  20.2. 

37  8»  From  the  preceding  illustrations  we  derive  the  following 
general 

RULE  FOR  MULTIPLYING  FEDERAL   MONEY. 

Multiply  as  in  simple  numbers,  and  point  Ojf  the  product  as  in 
multiplication  of  decimal  fractions.  (Art.  324.) 

OB3.  1 .  When  the  price,  or  the  quantity  contains  a  common  fraction,  the 
fraction  may  be  changed  to  a  decimal.  (Art.  337.) 

2.  In  business  operations,  when  the  mills  in  the  answer  are  5,  or  over,  it  is 
customary  to  call  them  a  cent ;  when  under  5,  they  are  disregarded. 

5.  What  cost  12£  yards  of  cotton,  at  9-J-  cts.  per  yard? 

Solution. — 12f  yards=12.5,  and  9i  cts.  =  .0925  ;  now  .0925  X 
125=$1. 15625.  Ans. 

6.  What  cost  45-J-  yards  of  satin,  at  87^  cts.  per  yard  ? 

7.  What  cost  169£  bbls.  of  pork,  at  $8|  per  barrel  ? 

8.  What  cost  324-f-  Ibs.  of  sugar,  at  12^  cts.  a  pound? 

9.  What  cost  97  gals,  of  oil,  at  87-£  cts.  per  gallon? 

10.  What  cost  310  Ibs.  of  tea,  at  62i  cts.  a  pound? 

11.  What  cost  23i  tons  of  hay,  at  $8$  per  ton  ? 

12.  What  cost  45  bbls.  of  flour,  at  $7|  per  barrel? 

13.  At  15i-  cts.  per  doz.,  what  cost  13£  dozen  of  eggs  ? 

14.  At  8f  cts.  per  pound,  what  will  32£  Ibs.  of  pork  come  to? 

15.  At  $6-}-  per  bbl,  what  will  145|  bbls.  of  flour  cost? 

16.  At  22-£  cts.  per  doz.,  what  will  a  gross  of  buttons  cost? 

17.  At  31^-  cts.  per  doz.,  what  cost  45  doz.  skeins  of  silk? 

18.  At  I7i  cts.  per  yard,  what  cost  91-£  yards  of  calico? 

19.  What  cost  45  doz.  plates,  at  62-£  cts.  per  doz.  ? 

20.  What  cost  63  doz.  pen-knives,  at  $3£  per  doz.  ? 

21.  What  cost  19  doz.  silver  spoons,  at  $7£  per  dozen? 

22.  What  cost  1865i  bushels  of  wheat,  at  $lf  per  bushel? 

23.  What  cost  2560^-  yds.  of  broadcloth,  at  $5|  per  yard? 

QUEST. — 378.  What  is  the  rule  for  Multiplication  of  Federal  Money?  Obs.  When  tJM 
wrice  or  quantity  contains  a  common  fraction,  what  should  be  done  with  it  ? 


234  DIVISION  OF  [SECT     X.L 

DIVISION   OP    FEDERAL   MONEY. 

Ex.  1.  A  man  bought  8  sheep  for  $42.24 :  what  did  n  give 
apiece  ? 

Analysis. — If  8  sheep  cost  $42.24,  1  sheep  will  cost  \  of  4^2.24 ; 
and  $42.24H-8  =  $5.28.  Am.  $5.28. 

PROOF. — If  1  sheep  costs  $5.25,  8  sheep  will  cost  8  times  as 
much,  and  $5.28 X8=$42. 24.  Hence, 

370.  When  the  number  of  articles,  pounds,  yards,  <fec.,  a. 
the  cost  of  the  whole  are  given,  to  find  the  price  of  one  article,  ont 
pound,  &c. 

Divide  the  whole  cost  by  the  whole  number  of  articles,  and  poin1 
off  the  quotient  as  in  division  of  decimal  fractions.  (Art.  330.) 

2.  A  shoemaker  sold  15  pair  of  boots  for  $67.50:  how  much 
did  he  get  a  pair  ? 

3.  A  merchant  sold  65£  Ibs.  of  sugar  for  $3.93:  how  much 
was  that  a  pound  ? 

4.  A  man  bought  6.5  yards  of  cloth  for  $20.345 :  how  much 
was  that  per  yard  ? 

5.  How  many  bbls.  of  flour,  at  $5.38  per  bbl.,  can  be  bought 
for  $34.97  ? 

Analysis. — Since  $5.38  will  buy  1  bbl.,  Operation. 

$34.97  will  buy  as  many  bbls.  as  $5.38  5.38)34.97(6.5  Ans 
are  contained  times  in  $34.97.     We  divide  32  28 

as  in  simple  numbers,  and  point  off  one  de-  2  699 

cimal  figure  in  the  quotient.  2  690 

PROOF. — $5. 38X6. 5 =$34.97,  the  given  amount. 

38O.  Hence,  when  the  price  of  one  article,  pound,  yard,  <fec., 
and  the  cost  of  the  whole  are  given,  to  find  the  number  of  arti- 
cles, &c. 

Divide  the  whole  cost  by  the  price  of  one,  and  point  off  the  quo* 
tient  as  in  division  of  decimals. 

GUEST.-  379.  When  the  number  of  articles,  pounds,  &c.,  and  the  cost  of  the  whole  are 
given,  how  is  the  cost  of  one  article  found  ?  380.  When  the  price  of  one  article,  one  pound* 
&.C.,  and  the  cost  of  the  whole  are  given,  how  is  the  number  of  articles  found  1 


ARTS.  379-381.]          FEDERAL  MONEY.  235 

6.  How  many  coats,  at  $12.56,  can  be  bought  for  $103.085  ? 

7.  How  many  times  is  $11.13  contained  in  87.606  ? 

8.  A  gentleman  distributed  $68  equally  among  32  poor  per- 
sons :  how  much  did  each  receive  ? 

Operation. 

32)£68(2.125  Ans.         After  dividing  the  $68  by  32,  there  is 

64  a  remainder  of  4  dollars,  which  should  be 

4000  reduced  to  cents  and  mills,  and  then  be 

32  divided  as  before.  (Art.  354.)    The  ciphers 

80  thus  annexed  must  be  regarded  as  deci- 

64  mals ;    consequently  there   will  be  three 

160  decimal  figures  in  the  quotient. 
160 
• 

381.  From  the  preceding  illustrations  we  derive  the  following 
general 

RULE   FOR   DIVIDING  FEDERAL   MONEY. 

Divide  as  in  simple  numbers,  and  point  off  the  quotient  as  in 
division  of  decimal  fractions.  (Art.  330.) 

OBS.  1.  In  dividing  Federal  Money,  if  the  number  of  decimals  in  the  divisor 
is  the  same  as  that  in  the  dividend,  the  quotient  will  be  a  whole  number. 
(Art.  330.  Obs.  1.) 

2.  When  there  are  mare  decimals  in  the  divisor  than  in  the  dividend,  annex 
as  many  ciphers  to  the  dividend  as  are  necessary  to  make  its  decimal  places 
equal  to  those  in  the  divisor.     The  quotient  thence  arising  will  be  a  whole 
number.  (Obs.  1.) 

3.  After  all  the  figures  of  the  dividend  are  divided,  if  there  is  a  remainder, 
ciphers  may  be  annexed  to  it,  and  the  operation  may  be  continued  as  in  divi- 
won  of  decimals.  (Art.  330.  Obs.  3.)     The  ciphers  thus  annexed  must  be  re- 
garded as  decimal  places  of  the  dividend. 

9.  How  many  gallons  of  molasses,  at  28  cts.  per  gallon,  can 
you  buy  for  886.25  ? 

QUEST.— 381.  What  is  the  rule  for  Division  of  Federal  Money  ?     Obs   When  there  is  a 
remainder  after  a  1  the  figures  of  the  dividend  are  divided,  how  proceed  ?    When  there  are 
more  decimals  in  the  divisor  than  in  the  dividend,  how  proceed! 
T.H. 


236  DIVISION  OF  [SECT.  XL 

10.  How  many  yards  of  calico,  at  13£  cts.  per  yard,  can  b« 
bought  for  $73.3  7£? 

11.  How  many  doz.  of  eggs,  at  9£  cts.  per  doz.,  can  be  bought 
for  $94.185? 

12.  At  18f  cts.  per  doz.,  how  many  skeins  of  sewing  silk  can 
be  bought  for  $67.50  ? 

13.  A  man  paid  $72.25  for  20.5  yards  of  cloth :  how  much  did 
he  pay  per  yard  ? 

14.  A  man  paid  $76.50  for  51  sheep :  what  was  the  price  per 
nead  ? 

15.  A  man  paid  $150  for  24  pair  of  boots:  how  much  was 
that  a  pair  ? 

16.  If  you  give  $56.25  for  28|  bbls.  of  flour,  how  much  do  you 
pay  per  barrel  ? 

17.  If  a  man  gives  $316.375  for  87^  yards  of  cloth,  what  is 
that  per  yard  ? 

18.  A  grocer  sold  965-£  Ibs.  of  sugar  for  $81.25  :  what  did  he 
get  a  pound  ? 

19.  The  fare  from  Albany  to  Buffalo,  a  distance  of  326  miles, 
is  $13.20  :  how  much  is  it  per  mile  ? 

20.  The  fare  from  Boston  to  Albany,  a  distance  of  203  miles 
is  $5.50  :  how  much  is  it  per  mile  ? 

21.  If  a  clerk's  salary  is  $650  per  annum,  how  much  does  he 
receive  per  day  ? 

22.  If  a  man  spends  $563.38  a  year,  how  much  are  his  average 
expenses  per  day  ? 

23.  At  87£  cts.  per  bushel,  how  many  bushels  of  wheat  can 
you  buy  for  $1500? 

24.  IIow  many  tons  of  coal,  at  $6.625  per  ton,  can  you  buy  for 
$752.36  ? 

25.  If  a  man's  income  is  $100  per  week,  how  much  is  it  per 
hour  ? 

26.  At  $14. 50  per  acre,  how  many  acres  of  land  can  you  buy 
foi  $3560  ? 

27.  At  $15^  apiece,  how  many  cows  can  you  buy  for  $7750  ? 

28.  At  $375.75  apiece,  how  many  carriages  cm  be  bought  for 
$56362.50  ? 


ART.  382.1  FEDERAL  MONET.  237 

COUNTING  ROOM  EXERCISES. 

Ex.  1.  What  cost  320  yards  of  satinet,  at  $1.12£  per  yard? 

Analysis. — If  the  price  were  $1  per  yard,  the  cloth  would  evi- 
dently cost  as  many  dollars  as  there  are  yards.  But  $1.12£  is 
equal  to  1  and  •£  dollars ;  hence,  the  cloth  will  cost  i  more  dollars 
than  there  are  yards ;  consequently,  if  we  add  to  the  number  of 
yards  i  of  itself,  it  will  give  the  cost.  Now  i  of  320=40,  and 
320+40=360.  Ans.  $360. 

PROOF. — $1.12£X320=$360,  the  same  as  before.     Hence, 

382.  When  the  price  of  1  article,  1  pound,  &c.,  is  $1.12£ 
$1.25  ;  $1.37£;  &c.,  to  find  the  cost  of  any  number  of  articles. 

To  the  given  number  of  articles,  add  \,  \,  -f ,  &c.,  of  itself,  as 
the  case  may  be,  and  the  sum  will  be  the  cost  required. 

OBS.  When  the  price  of  1  article,  &c.,  is  $2.12$,  $2.25,  $3.37j,  &<;.,  the 
operation  may  be  contracted  by  multiplying  the  given  number  of  articles  by 
^s,  2},  3f ,  &c.,  as  the  case  may  be. 

2.  What  cost  640  bushels  of  wheat,  at  $1.25  per  bushel? 

3.  What  cost  372  pair  of  shoes,  at  $1.37^  a  pair? 

4.  What  cost  480  bbls.  of. cider,  at  $1.62£  a  barrel? 

5.  What  cost  520  yards  of  silk,  at  $1.50  per  yard? 

6.  What  cost  720  drums  of  figs,  at  $1.87-£  per  drum? 

7.  At  $2.12-£  apiece,  what  will  480  sheep  cost? 

8.  _At  $2.37-£  apiece,  what  will  364  vests  cost? 

9.  At  $3.25  per  yard,  what  cost  744  yards  of  cloth? 

10.  At  $4.62i  apiece,  what  cost  960  hats? 

11.  At  $5.12£  a  pair,  what  cost  278  pair  of  boots? 

12    At  $7.37^  per  lb.,  what  will  365  Ibs.  of  opium  cost? 

13.  A  collier  sold  856  tons  of  coal,  at  $6.87i  per  ton:  how 
much  did  it  amount  to  ? 

14.  At  19. 62^  per  acre,  what  will  537  acres  of  land  cost? 
lo.  What  cost  72  Ibs.  of  flax,  at  $8.25  per  hundred? 

Analysis. — 72  pounds  are  iW  of  100  pounds;   th»;r«eiOre  72 

Q  2{Sv72 
pounds  will  cost  flft  of  $8.25  ;  and  -fifo  of  $8.25=—  ~-. 


238  APPLICATIONS    OF  [SECT.   XI. 

Operation.  We  multiply  the  price  of  100  Ibs.  (88.25) 

$8.25  by  72,  the  given  number  of  pounds,  and  the 

72  product  $594.00,   is  the  cost  of  72   Ibs.  at 

1050  $8.25  per  pound.     But  the  price  is  $8.25  per 

5775  hundred;  consequently  the  product  $504.00 

$5.9400  Ans.       is  100  times  too  large,  and  must  therefore  be 

divided  by  100,  to  give  the  true  answer.    But 

to  divide  by  100,  we  simply  remove  the  decimal  point  two  places 

towards  the  left.  (Art.  331.) 

16.  What  cost  367  bricks,  at  $4.45  per  1000? 

Operation.  Reasoning  as  before,  367   bricks  will  cost 

4.45  T^Vs-  of  $4.45.    We  multiply  the  price  of  1000 

3  67  bricks  by  the  given  number  of  bricks,  and  di- 

$1.63315  Ans.  vide  the  product  by  1000.  (Art.  331.)    Hence, 

383.  To  find  the  cost  of  articles  sold  by. the  100,  or  1000. 

Multiply  the  given  price  by  the  given  number  of  articles ;  then 
if  the  price  is  for  100,  divide  the  product  by  100  ;  but  if  the  price 
is  for  1000,  divide  it  by  1000.  (Art.  331.) 

17.  A  farmer  sold  563  Ibs.  of  hay,  at  $1.12£  per  hundred  :  how 
much  did  it  come  to  ? 

18.  What  cost  1640  Ibs.  of  beef,  at  $6.37i  per  hundred? 

19.  What  cost  2719  Ibs.  of  fish,  at  $4.20  per  hundred? 

20.  What  is  the  freight  on  3568  Ibs.  from  New  York  to  Buffalo, 
ai$1.67  per  hundred? 

21.  What  cost  6521  Ibs.  of  cheese,  at  7f  cts.  per  hundred? 

22.  What  cost  15214  Ibs.  of  butter,  at  12|  cts.  per  hundred? 

23.  At  $6.25  per  1000,  what  cost  865  feet  of  spruce  boards? 

24.  At  $19.45  per  1000,  what  cost  2680  feet  of  pine  boards? 
25     At  $67.33  per  1000,  what  cost  6500  feet  of  mahogany? 
26.  When  ginger  is  §16.53  per  cwt.,  what  is  it  per  pound? 

Analysis. — Since  100  Ibs.  cost  $16.53,  1  Ib.  will  cost  -rbr  of' 
I '6.53.  But  to  divide  by  100,  we  remove  the  decimal  point  twa 
places  to  the  left.  (Art.  331.)  Ans.  $0  1653. 

Q,OBS  r.— 383.  How  do  you  find  the  cost  of  articles  sold,  by  the  100,  or  1000  7 


ARTS.  383-385.]  FEDERAL  MONEY  239 

27.  When  pine  boards  are  $21.63  per  1000,  what  are  they  per 
foot? 

Solution. — Reasoning  as  before,  1  foot  will  cost  -reVv  of  $21.63. 
Now  to  divide  by  1000,  we  remove  the  decimal  point  three  places 
to  the  left.  (Art.  331.)  Ans.  $.02163.  Hence, 

384.  "When  the  cost  of  100,  or  1000  articles,  pounds,  &c.,  4 
given,  the  price  of  one  is  found  by  simply  removing  the  decimal 
point  in  the  given  cost  or  dividend,  as  many  places  to  the  left  as 
there  are  ciphers  in  the  divisor.  (Art.  331.) 

28.  Bought  1000  bricks  for  $7.20  :  what  is  that  apiece? 

29.  If  1000  feet  of  hemlock  boards  cost  $6.40,  what' will  one 
foot  cost  ? 

30.  Bought  42  cwt.  of  tobacco  for  $565.82  :  what  is  that  per 
cwt. ;  and  what  per  pound  ? 

31.  Bought  75  cwt.  of  butter  for  $966.38:  what  is  that  per 
cwt. ;  and  what  per  pound  ? 

BILLS,    ACCOUNTS,    <feC. 

385*  A  Bill,  in  mercantile  operations,  is  a  paper  containing 
a  written  statement  of  the  items,  and  the  price  or  amount  of  goods 
sold. 

32.  What  is  the  cost  of  the  several  articles,  and  what  the  amount, 
of  the  following  bill  ? 

NEW  YORK,  May  21st,  1847. 
G.  B.  Grannis,  Esq., 

Bought  of  Mark  H.  Newman  &  Co.9 
75  Thomson's  Mental  Arithmetic,     at         $  .12£ 
50         "  Practical  Arithmetic,  "  .31 1 

36  Porter's  Rhetorical  Reader,  '«  .62£ 

25  Willson's  School  History,  "  .46  -         -    ' 

80  M'Elligott's  Young  Analyzer,       "  .31-J- 

75  Thomson's  Day's  Algebra,  "  .50  -         • 

60         "  Lcgendre's  Geometry,  .47£         -         • 

Received  Payment, 

Mark  fl.  Newman  <&  Co. 


240  BILLS,    ACCOUNTS,    ETC.  [SECT.  XI. 

(33.) 

PHILADELPHIA,  June  10th,  1847. 
Hon.  Horace  Binney, 

Sought  of  Leverette  &  Griggst 
1G3  Ibs.  Butter,  at         $  .14£ 

235  Ibs.  Coffee,  "  .08* 

86  Ibs.  Chocolate,          "  .11 

685  Ibs.  Sugar,  "  ,10|- 

21  doz.  Eggs,  "  .13 

860  Ibs.  Lard,  "  .09|- 

What  was  the  cost  of  the  several  articles,  and  what  the  amount 
of  his  bill? 

(84.) 

ALBANY,  July  1st,  1847. 
Messrs.  Collins  &  Brotliers, 

To  G.  W.  Bunker,  Dr. 
For  320  yds.  Silk,  at         $1.12£ 

"    256    "     Broadcloth,      "  3.62£ 

"    175  pair  Cotton  Hose,  ".  0.12£ 

"    100    "     Silk          "       "  0.871 

"      15  doz.  Gloves,  "  0.621 

"    120  Straw  Hats,  "  1.871 

What  was  the  cost  of  the  several  articles,  and  what  the  amount 
of  his  bill? 

(35.) 

ST.  Louis,  Aug.  25th,  1847. 
James  Henry,  Esq. 

To  J.  L.  Hoffman  &  Co.,  Dr. 

For  15260  Ibs.  Pork,  at         $0.051 

"      7265  Ibs.  Cheese,         "  0.081 

"    11521  bu.  Corn,  "  0.50 

"      1560  bbls.  Flour,          "  6.12^ 

CREDIT. 

Bj  115C  Ibs,  Cotton,  at         80.06| 

"    8256  Ibs.  Sugar,  «  0.07 

M    6450  gals.  Molasses,        "  0.3 7i 

"    Cash  to  balance  account,  -  - 

What  is  the  amount  of  cash  requisite  to  balance  the  account  ? 


ARTS.  386,  387. j  PERCENTAGE.  241 

SECTION    XII. 
PERCENTAGE. 

ART   386*  The  terms  Percentage  and  Per  Cent,  signify  a  cer- 
tain allowance  on  a  hundred;  that  is,  a  certain  part  of  a  hundred, 
or  simply  hundred tks.     Thus,  the  expression  G  per  cent,  signifies 
6  hundredths,  (ruir,)  V  per  cent.,  7  hundredths,  (-rfcr,)  &c.,  of  the  " 
number,  or  sum  of  money  under  consideration. 

Note.—  -The  terms  Percentage  and  Per  Cent,  are  derived  from  the  Latin  per 
and  centum,  signifying  by  I/M  hundred. 

387.  We  have  seen  that  hundredth*  are  decimal  expressions, 
occupying  the  first  two  places  of  figures  on  the  right  of  the  deci- 
mal point.  (Arts.  311,  314.)  Now,  since  percentage  and.  per  cent. 
signify  hundredths,  it  is  manifest  that  they  can  be  expressed  by 
decimals,  as  in  the  following 

PERCENTAGE    TABLE. 

1  per  cent.  .  .  .  is  written  thus :  .01 

2  per  cent.  .  "        "         "  .02 

3  per  cent.            ...              "        "        "  .03 
G  per  cent.            .            .            .              "        "        "  .00 
7  per  cent             ...              u        u        u  .07 

10  per  cent.             .             .             .  «  «  «  JO 

12  per  cent.            .    *                    .  "  «  "  ,13 

50  per  cent.             .             .             .  «  «  «  .50 

100  per  cent.             .             .  "  "  "  1.00 

103  per  cent.             .             .             .  «  «  »  1.03 

1-25  per  cent.,  &c.                 .            .  «  «  «  1/25 

J  per  cent.,  that  is,  $  of  1  per  cent  "  "  "  .005 

\  per  cent.,  that  is,  \  of  1  per  cent.  "  "  "  0025 

|  per  cent.,  that  is,  $  of  1  per  cent.  "  «  "  .0075 

13 1  per  cent.             ...  "  "  "  .13125       jjt 

251  per  cent.             .             .             .  «  ,c  «  .25375 

OBS.  I.  It  will  be  seen  from  the  preceding  Table,  that  when  the  given  per 
cent,  is  /(.'.s.s  than  10,  a  cipher  must  be  prefixed  to  the  ligure  ex  pressing  it,  in  the 
same  manner  as  when  the  number  of  cents  is  less  than  10.  (Art.  3GU.  Obs.  1.) 

QPKST. — 386.  Wh;vt  do  the  terms  percentage  and  per  cent,  signify  1  What  is  meant  IT 
6  oer  cent.,  7  per  cent.,  &.C.,  of  any  number,  or  sum  1 


242  PERCENTAGE.  [SECT.   Xll. 

When  the  given  per  cent,  is  more  than  100,  it  must  plainly  require  a  mixed 
number  to  express  it.  (Art.  315.  Obs.  2.) 

2.  Parts  of  1  per  cent,  may  be  expressed  either  by  a  common  fraction,  or  by 
decimals      Thus,  the  expression  17f  per  cent.,  is  equivalent  to  .17G25  percent. 

3.  The  first  two  decimal  figures  properly  denote  the  per  cent.,  for  they  are 
kundredtks ;  the  other  decimals  being  parts  of  hundrcdths,  express  parts  of 
1  per  cent. 

EXAMPLES. 

1  Write  1  per  cent.,  2  per  cent.,  4  per  cent.,  6  per  cent.,  7  per 
cent.,  8  per  cent.,  in  decimals. 

2  Write  11  per  cent. ;  12;  14;  15;  16;  23;  65;  93. 

3  Write  i  per  ct. ;  i;i;|;  f;  *;  i;  i;  f;  f ;  1;  *;  };  *. 

4.  Write  4i  per  ct. ;  6|;  ?i;  9i;  12|;  16-J-;  115;  400|. 

5.  An  agent  collected  $700  for  a  merchant,  and  received  5  per 
cent,  for  his  services :  how  much  did  he  receive  ? 

Analysis. — Since  5  per  cent,  is  the  same  as  -p^,  the  agent  must 
have  received  -Hhr  of  $700.  Now  rb-  of  $700  is  $f  JHK  which  is 
equal  to  $7 ;  and  5  hundredths  is  5. times  $7,  or  $35. 

Operation.  Since  TOTT— -05,  we  multiply  the  given  num- 

$700  ber  of  dollars  by  .05,  and  it  gives  the  answer  in 

.05    •  cents,  which  we  reduce  to  dollars  by  pointing 

$35TOO  Ans.  off  2  decimals.  (Art.  372.)     Hence, 

388.  To  calculate  percentage  on  any  number,  or  sum  of 
money.  ^ 

Multiply  the  given  number  or  sum  by  the  given  per  cent,  expressed 
decimally ;  and  point  off  ike  product  as  in  multiplication  of  deci- 
mal fractions.  (Art.  324.) 

OBS.  1.  It  is  important  for  the  learner  to  observe,  that  the  amount  of  money 
collected,  is  made  the  basis  upon  which  the  percentage  is  computed.  That  is, 
the  agent  is  entitled  to  3  dollars,  as  often  as  he  collects  100  dollars,  and  not  as 
n  as  he  pays  over  100  dollars,  as  is  frequently  supposed.  For  in  the  latter 
case  he  would  receive  only  -f-f  3,  instead  of  -,-•§•  -„-  of  the  sum  in  question.  This 
distinction  is  important,  especially  in  calculating  percentage  on  large  sums. 

QUEST. — 387.  How  may  per  centage  or  per  rent,  be  expressed?  Obs.  When  the  given 
per  cent,  is  less  than  10,  how  is  it  written?  When  more  than  100,  how?  388.  Hew  is 
percentage  calculated  ?  Obs.  In  collecting  money,  upon  what  basis  is  the  per  cent,  cal- 
culated ?  If  the  per  cent,  contains  a  common  fraction  which  cannot  le  expressed  deci- 
mally, how  proceed  1 


ART.  388  ]  PERCENTAGE.  243 

2.  Hence,  if  the  per  cent,  contains  a  common  fraction  which  cannot  be  ex- 
pressed decimally,  first  multiply  by  the  decimal,  then  by  the  common  fraction 
of  the  given  per  cent.,  and  point  off  the  sum  of  their  products  as  above. 

6.  What  is  4|  per  cent,  of  $300  ? 

Solution. — Expressed  decimally,  4£  per  cent.=.042  ;  (Art.  387 
Obs.  2  ;)  and  $300  X  .042-$! 2.60.  4ns. 

7.  What  is  3  per  cent,  of  $256.25  ? 

8.  What  is  2  per  cent,  of  $437.63  ? 

9.  What  is  2\  per  cent,  of  $138.432  ? 
10  What  is  6  per  cent,  of  $145.13  ? 

11.  What  is  7  per  cent,  of  $1630.10? 

12.  A  man  borrowed  $150,  and  paid  7  per  cent,  for  the  use  or 
it :  how  much  did  he  pay  ? 

13.  A  merchant  bought  goods  amounting  to  $1825,  and  sold 
them  so  that  he  gained  12  per  cent. :  how  much  did  he  gain? 

14.  A  constable  collected  $862.56,  and  charged  5  per  cent',  for 
his  services :  how  much  did  he  receive ;  and  how  much  did  he 
pay  over  ? 

15.  What  is  10  per  cent,  of  $4020.50  ? 

16.  What  is  8  per  cent,  of  $1675  ? 

17.  What  is  4£  per  cent,  of  $725  ? 

18.  What  is  5|  per  cent,  of  $648.30? 

19.  What  is  6|  per  cent,  of  $1000  ? 

20.  WThat  is  7i  per  cent,  of  $2000  ? 

21.  What  is  8|  per  cent,  of  $100.25  ? 

22.  A  farmer  having  1500  sheep,  lost  25  per  cent,  of  them: 
how  many  did  he  lose  ? 

23.  A  merchant  having  $1960  on  deposit,  drew  out  20  per  cent, 
of  it :  how  much  had  he  left  in  the  bank  ? 

24.  A  merchant  imported  1500  boxes  of  oranges,  and  12£  per 
cent,  of  them  decayed :  how  many  boxes  did  he  lose ;  and  how 
many  had  he  left  ? 

25..  What  is  £  per  cent  of  $1625  ? 

26.  What  is  i  per  cent,  of  $2526.40  ? 

27.  What  is  i  per  cent,  of  $42260.08  ? 

28.  What  is  i  per  cent,  of  $75000  ? 

29.  What  is  -f  pvr  cent,  of  $100000? 

11* 


244  PERCENTAGE.  [SECT.  XII. 

30.  What  is  i  per  cent,  of  $45241.20  ? 

31.  What  is  i  per  cent,  of  $675264  ? 

32.  A  merchant  bought  a  stock  of  goods  amounting  to  $4565, 
and  paid  3£  per  cent,  for  freight :  what  was  the  whole  cost  of 
his  goods  ? 

33.  A  man's  salary  is  $2000  a  year,  and  he  lays  up  37£  pel 
cent,  of  it :  how  much  does  he  spend  ? 

34.  A  youth  who  inherited  $20000,  spent  40  per  cent,  of  it  in 
dissipation :  how  much  had  he  left  ? 

35.  Two  merchants  embarked  in  business  with  $18250  capital 
apiece  ;  one  gained  20  per  cent,  and  the  other  lost  20  per  cent, 
the  first  year :  what  was  then  the  amount  of  each  man's  property  ? 

36.  Two  men  invested  -Si 0000  apiece  in  stocks;  one  lost  8  per 
cent.,  the  other  6  per  cent. :  what  was  the  difference  of  their  loss  ? 

37.  What  is  the  difference  between  6  per  cent,  of  $1040,  and 
7  per  cent,  of  $905  ? 

.     APPLICATIONS   OP  PERCENTAGE. 

389»  PERCENTAGE,  or  the  method  of  reckoning  by  hundred ths, 
is  applied  to  various  calculations  in  the  practical  concerns  of  life. 
Among  the  most  important  of  these  are  Commission,  Brokerage, 
the  Rise  and  Fall  of  Stocks,  Interest,  Discount,  Insurance,^  Profit 
and  Loss,  Duties,  and  Taxes.  Its  principles,  therefore,  should  be 
thoroughly  understood  by  every  scholar. 

COMMISSION,   BROKERAGE,  AND   STOCKS. 

39O«  Commission  is  the  per  cent,  or  sum  charged  by  agents 
for  their  services  in  buying  and  selling  goods,  or  transacting  other 
business. 

OBS.  An  Agent  who  buys  and  sells  goods  for  another,  is  called  a  C(nnmi9* 
sion  Merchant,  a  Factor,  or  Correspondent. 

391.  Brokerage  is  the  per  cent,  or  sum  charged  by  money  deal- 
ers, called  Brokers,  for  negotiating  Bills  of  Exchange,  and  other 
monetary  operations,  and  is  of  the  same  nature  as  Commission. 

Q.U«ST.— 390.  What  is  commission  1  Obs.  What  is  au  agent  who  buys  and  sells  goad* 
for  another  usually  called  ? 


ARTS.  389-395.]  COMMISSION.  245 

392.  By  the  term  Stocks,  is  meant  the  Capital  of  moneyed 
institutions,  as  incorporated  Banks,  Manufactories,  Railroad  and 
Insurance  Compiinies ;  also,  Government  and  State  Bonds,  &c. 

OBS.  1.  Stocks  are  usually  divided  into  portions  of  $100  each,  called  shares; 
anil  the  owners  of  these  shares  are  culled  Stockholders. 

2.  The  association  or  company  thus  formed,  is  called  a  corporations  the  in- 
strument specifying  the  powers,  rights,  and  privileges  invested  in  the  corpora- 
tion /  is  called  a  charter. 

393.  The  original  cost  or  valuation  of  a  share  is  called  its 
nominal,  or  par  value  /  the  sum  for  which  it  can  be  sold,  is  its 
real  value. 

OES.  1.  The  rise  or  fall  of  Stocks  is  reckoned  at  a  certain  per  cent,  of  its 
par  value.  The  term  par  is  a  Latin  word,  which  signifies  equal,  or  a  stale  of 
equality. 

2.  When  Stocks  sell  for  their  original  cost  or  valuation,  they  are  said  to  be 
at  par ;  when  they  sell  for  more  than  cost,  they  are  said  to  be  above  par,  at  a 
premium,  or  an  advance ;  when  they  do  not  sell  at  cost,  they  are  said  to  be 
below  par,  or  at  a  discount. 

3.  Persons  who  deal  in  Stocks  are  usually  called  Stock  Brokers,  or  Stock 
Jobbers. 

394.  The  commission  or  allowance  made  to  factors  and  brokers, 
also  the  rise  and  fall  of  stocks,  are  usually  reckoned  at  a  certain 
vercentaye  on  the  amount  of  money  employed  in  the  transaction, 
or  on  the  par  value  of  the  given  shares.     Hence, 

395*  To  compute  commission,  brokerage,  and  the  premium  or 
discount  on  stocks. 

Multiply  the  given  sum  l>y  the  given  per  cent,  expressed  in  deci- 
mal*, and  point  off  the  product  as  in  Percentage.  (Art.  388.) 

OBS.  The  commission  for  the  collection  of  bills,  taxes,  &c.,  also  for  the  sale 
or  purchase  of  goods,  varies  from  *2i  to  12  or  15  per  cent.,  and  should  always 
be  reckoned  on  the  amount  of  money  collected,  or  paid  out.  or  employed  in  the. 
transaction. 

The  brokerage  for  the  sale  or  purchase  of  stocks,  varies  from  J  tr  f  pe 
cent.,  reckoned  on  the  par  value  of  the  stock. 

QUEST.— :<91.  What  is  hrnkernpe  1  :?92.  Whnt  is  meant  hy  the  term  storks  1  Ob.-  How 
are  stocks  nnially  «fivi<1pd  1  .'!<):{.  What  is  the  par  value  of  stocks  ?  What  the  real  value  ? 
Obs.  What  is  the  meaning  of  the  term  par  ?  When  are  stocks  at  par  ?  When  above  par! 
When  below  1  395.  How  do  you  compute  commission,  brokerage,  &.c.  t 


246  COMMISSION.  JSECT.  XII 

EXAMPLES. 

1.  An  auctioneer  sold  goods  amounting  to  $463,  at  3  per  cent 
commission:  how  much  did  he  receive ?     Am.  $13.89. 

2.  An  agent  bought  goods  amounting  to  $625.375  :  what  is 
his  commission,  at  2  per  cent.  ? 

3.  What  is  the  commission  on  $1682.25,  at  3-£  per  cent.  ? 

4.  What  is  the  commission  on  $1463.18,  at  5  per  cent.  ? 

5.  What  is  the  commission  on  $2560.07,  at  4£  per  cent.  ? 

6.  What  is  the  commission  on  $10250,  at  6  per  cent.  ? 

7.  What  is  the  commission  on  $8340.60,  at  7  per  cent.  ? 

8.  What  is  the  commission  on  $960.625,  at  5i  per  cent.  ? 

9.  A  commission  merchant  sold  goods  to  the  amount  of  $6235, 
at  2£  per  cent. :  what  was  his  commission  ? 

10.  An  attorney  collected  a  debt  of  $8265.17,  and  charged  7£ 
per  cent,  for  his  services :  how  much  did  he  receive  ? 

11.  Bought  $1108  worth  of  books,  at  4  per  cent,  commission: 
what  was  the  amount  of  commission  ? 

12.  A  tax-gatherer  collected  $12250,  for  which  he  was  entitled 
to  5i  per  cent,  commission :  how  much  did  he  receive  ? 

13.  Sold  goods  amounting  to  $1432.26:   how  much  was  the 
commission,  at  4  per  cent.  ? 

1 4.  A  commission  merchant  sold  a  quantity  of  hardware  amount- 
ing to  $9240.71  :   how  much  would  he  receive,  allowing  2^  per 
cent,  for  selling,  and  2  per  cent,  more  for  guaranteeing  the  pay- 
ment? 

15.  An  auctioneer  sold  carpeting  amounting  to  $2136.63,  and 
charged  2^  per  cent,  for  selling,  and  2f  per  cent,  for  guaranteeing 
the  payment:  how  much  did  the  auctioneer  receive;   and  how 
much  did  he  remit  the  owner  ? 

396.  Commission  merchants,  agents,  &c.,  generally  keep  an 
account  with  their  employers,  and  as  they  make  ii,  vestments  or 
sales  of  goods,  charge  their  commission  on  the  amount  invested, 
or  the  sum  employed  in  the  transaction. 

Sometimes,  however,  a  specific  amount  is  sent  to  an  agent  or 
broker,  requesting  him,  after  deducting  his  comnrSsion,  to  lay  out 
the  balance  in  a  certain  manner. 


ARTS.  396,  397.]  BROKERAGE.  247 

16.  A  gentleman  sent  his  'agent  $1500  to  purchase  a  library: 
how  much  had  he  to  lay  out  after  deducting  his  commission  at  5 
per  cent. ;  and  what  was  his  commission  ? 

Note. — The  money  actually  laid  out  by  the  agent  in  books,  is  manifestly  the 
proper  basis  on  which  to  calculate  his  commission  ;  for  it  would  be  unjust  tt 
charge  commission  on  the  sum  he  retains.  (Art.  395.  Obs.) 

Analysis. — The  money  laid  out  is  •}•$•$  of  itself,  and  the  commis- 
sion is  -poTT  °f  this  sum ;  consequently  the  money  laid  out  added 
to  the  commission,  must  be  -HHj*  the  whole  amount.  The  question 
therefore  resolves  itself  into  this:  $1500  is  ^-{J-ft  of  what  sum? 
If  $1500  is  -HHK  rh-  must  be  1500-j-lOo^W,  and  1^=-^° 
X  100  —  $1428.57,  the  sum  laid  out.  Now  $1500 — $1428.57  — 
$71.43,  the  commission. 

PROOF.—- $1428.57 X. 05=171.43;  and  $1428.57+$7l.43  = 
$1500,  the  amount  sent.  Hence, 

397.  To  compute  commission  when  it  is  to  be  deducted  in 
advance  from  a  given  amount,  and  the  balance  is  to  be  invested. 

Divide  the  given  amount  by  $1  increased  by  the  per  cent,  commis- 
sion, and  the  quotient  will  be  the  part  to  be  invested.  Subtract  the 
part  invested  from  the  given  amount,  and  the  remainder  will  be  the 
commission. 

OBS.  The  commission  may  also  be  found  by  multiplying  the  sum  invested  by 
the  given  per  cent,  according  to  the  preceding  rule.  (Art.  395.) 

17.  An  8q;ent  received  $21500  to  lay  out  in  provisions,  after 
deducting  2  A)er  cent,  commission :  Avhat  sum  did  he  lay  out  ? 

18.  A  country  merchant  sent  $3560  to  his  agent  in  the  city,  to 
purchase,  goods :  after  taking  out  his  commission,  at  3£  per  cent., 
how  much  remained  to  lay  out  ? 

19.  Baring,  Brothers  &  Co.  sent  their  agents  $800000  to  buy 
flour :  after  deducting  5  per  cent,  commission,  how  much  would 
be  left  to  invest? 

20.  A  broker  negotiated  a  bill  of  exchange  of  $82531,  at  5  per 
cent. :  how  much  did  he  receive  for  his  services  ? 

21.  What  is  the  brokerage  on  $94265,  at  1|  per  cent.? 

22.  What  is  the  brokerage  on  $6200,  at  £  per  cent.  ? 


248  STOCKS.  [SECT.  XII 

23.  What  is  the  brokerage  on  $8845.50,  at  £  per  cent.  ? 

24.  What  is  the  brokerage  on  $2500,  at  •£  per  cent.  ? 

25.  A  broker  made  an  investment  of  $21265,  and  charged  1^ 
per  cent. :  what  was  the  amount  of  his  brokerage  ? 

20.  If  you  buy  20  shares  of  Western  Railroad  stock,  at  7  per 
cent,  advance,  how  much  will  your  stock  cost  you?  Ans.   $2140. 

Note. — The  stock  evidently  cost  its  par  value,  which  is  $'2000  and  7  per  cent. 
tesides.    Now  $2000X.07=$140.00;  and  $2000 -f-$140=$2140. 

27.  What  cost  20  shares  of  bank  stock,  at  7  per  cent,  discount? 

Ans.  $2000— $140  =  $]  860. 

28.  What  cost  35  shares  of  New  York  and  Erie  Railroad  stock, 
at  5-£  per  cent,  premium  ? 

29.  A  merchant  bought  45  shares  of  Commercial  Bank  stock, 
at  par,  and  afterwards  sold  them,  at  50  per  cent,  discount :  how 
much  did  he  lose  ? 

30.  A  man  invested  $8460  in  the  New  England  Manufacturing 
Co.,  and  afterwards  sold  out  at  4£  per  cent,  advance :  how  much 
did  he  sell  his  stock  for  ? 

31.  Sold  64  shares  of  Hudson  River  Railroad  stock,  at  10£  per 
cent,  premium  :  how  much  did  they  come  to  ? 

32.  A  man  bought  35  shares  of  Utica  and  Syracuse  Railroad 
stock,  at  par,  and  afterwards  sold  them  at  l£  per  cent,  advance : 
how  much  did  he  get  for  them  ? 

33.  A  man  bought  15  shares  of  Albany  and  Schenectady  Rail- 
road stock,  at  2  per  cent,  advance,  and  sold  them  at  10  per  cent, 
disc. :  how  much  did  he  sell  them  for ;  and  how  much  did  he  lose  ? 

34.  Bought  71  shares  in  the  Albany  Gas  Co.  at  5-£  per  cent. 
'  premium  :  how  much  did  they  amount  to  ? 

35.  A  broker  bought  48  shares  of  Michigan  Railroad  stock, 
at.  14  per  cent,  discount,  and  sold  them  at  6  per  cent,  advance: 
how  much  did  he  make  by  the  operation  ? 

36.  If  I  employ  a  broker  to  buy  me  55  shares  of  Railroad  stock, 
rhich  is  20  per  cent,  below  par,  and  pay  him  £  per  cent,  broker 
age,  how  much  will  my  stock  cost  me  ? 

37.  If  my  agent  buys  78  shares  of  New  York  and  PhihiJel- 
phia  Railroad  stock,  at  15  per  cent,  advance,  and  charges  me  -f 
per  cent,  brokerage,  how  much  will  rnj  stock  cost? 


ARTS.  398-40 l.J  INTEREST.  249 

INTEREST. 

398.  INTEREST  is  the  sum  paid  for  the  use  of  money  by  the 
borrower  to  the  lender.     It  is  reckoned  at  a  given  per  cent,  per 
annum ;  that  is,  so  many  dollars  are  paid  for  the  use  of  $100 
for  one  year  ;  so  many  cents  for  100  cents;  so  many  pounds  for 
£100;  <fec. 

OBS.  The  student  should  be  careful  to  notice  the  distinction  between  Com- 
mission, and  Interest.  The  former  is  reckoned  at  a  certain  per  cent,  without 
regard  to  time;  (Art.  395;)  the  latter  is  reckoned  at  a  certain  percent,  for  one 
year ;  consequently,  for  longer  or  shorter  periods  than  one  year,  like  proportions 
of  the  percentage  for  one  year  are  taken. 

The  term  per  annum,  signifies  for  a  year. 

399.  The  money  lent,  or  that  for  which  interest  is  paid,  is 
called  the  principal. 

The  per  cent,  paid  per  annum,  is  called  the  rate. 

The  sum  of  the  principal  and  interest,  is  called  the  amount. 
Thus,  if  1  borrow  $100  for  1  year,  and  agree  to  pay  5  per  cent, 
for  the  use  of  it,  at  the  end  of  the  year  I  must  pay  the  lender 
$100,  the  sum  which  I  borrowed,  and  $5  interest,  making  $105. 
The  principal  in  this  case,  is  $100;  the  interest  $5;  the  rate  5 
per  cent. ;  and  the  amount  $105. 

OBS.  The  term  per  annum,  is  seldom  expressed  in  connection  with  the  rale 
per  u'.'if.,  but  it  is  always  understood ;  for  the  rate  is  the  per  cent,  paid  per 
annum.  (Art.  399.) 

400.  The  rate  of  interest  is  usually  established  by  law.     It  va- 
nes in  different  countries  and  in  different  parts  of  our  own  country. 

OBS.  When  no  rate  is  mentioned,  the  rate  established  by  the  laws  of  the 
State  in  which  the  transaction  takes  place,  is  always  understood  to  be  the  one 
intended  by  the  parties. 

40 1 .  Any  rate  of  interest  higher  than  the  legal  rate,  is  called 
usury,  and  the  person  exacting  it  is  liable  to  a  heavy  penalty. 

Any  rate  less  than  the  legal  rate  may  be  taken,  if  tht  parties 
concerned  so  agree. 


QUKST. — 398.  What  is  Imerest?  How  is  it  reckoned?  Obs.  What  is  the  difference  be- 
tween Commission  and  Interest  ?  What  is  meant  by  the  term  per  annum  7  3!>9.  What  is 
nteant  by  the  principal?  The  rate  ?  The  amount  ?  409.  How  is  the  rate  usually  deter- 
mined ?  Is  it  the  same  everywhere  ?  Obs.  When  no  rate  is  mentioned,  what  rate  is  un- 
derstood 7  401.  What  is  anv  rate  higher  than  the  legal  rate  called  ? 


250 


INTEREST. 


[SECT.  XII. 


4O2»  The  legal  rates  of  interest,  and  the  penalty  for  usury  in 
the  several  States  of  the  Union,  are  as  follows : 


Penalty  for  Usury. 
Forfeit  of  the  whole  debt. 
Forfeit  of  three  times  the  usury. 
Recovery  in  action  with  costs. 
Forfeit  of  three  times  the  usury. 
Forfeit  of  the  usury  and  int.  on  the  debt. 
Forfeit  of  the  whole  debt. 
Forfeit  of  the  whole  debt. 
Forfeit  of  the  whole  debt. 
Forfeit  of  the  whole  debt. 
Forfeit  of  the  whole  debt. 
Usurious  contracts  void. 
Forfeit  of  double  the  usury. 
Forfeit  of  double  the  usury. 
Forfeit  of  interest  and  usury  with  costs. 
Forfeit  of  three  times  the  usury. 
Forfeit  of  interest  and  usury. 
Forfeit  of  usury  and  costs. 
Usurious  contracts  void. 
Usurious  contracts  void. 
Usury  may  be  recovered  with  costs. 
Usuiious  contracts  void. 
Forfeit  of  double  the  excess. 
Forfeit  of  three  times  the  usury,  and  int.  due. 
Forfeit  of  the  usury,  and  the  interest  due. 
Forfeit  of  the  usury,  and  one  fourth  the  debt. 
Forfeit  of  usury. 
Forfeit  of  interest  and  usury. 
Forfeit  of  three  times  the  usury. 
Forfeit  of  three  times  the  usury. 
Usurious  contracts  void. 
Usurious  contracts  void. 


OBS.  1.  On  debts  and  judgments  in  favor  of  the  United  States,  interest  M 
computed  at  6  per  cent. 

2.  In  Canada  and  Nova  Scotia,  the  legal  rate  of  interest  is  6  per  cent.  In 
England  and  France  it  is  5  per  cent. ;  in  Ireland  6  per  cent.  In  Italy,  about 
the  commencement  of  the  13th  century,  it  varied  from  20  to  30  per  cent. 


States. 

Legal  rates. 

Maine, 

6  per  cent. 

N.  Hampshire, 

6  per  cent. 

Veimont, 

6  per  cent. 

M  assachusetts, 

6  per  cent. 

Rhode  Island, 

6  per  cent. 

Connecticu  , 

6  per  cent. 

New  York, 

7  per  cent. 

New  Jersey, 

6  per  cent. 

Pennsylvania, 

6  per  cent. 

Delaware, 

6  per  cent. 

Maryland, 

6  per  cent,  a 

Virginia, 

6  per  cent. 

N.  Carolina, 

6  per  cent. 

S.  Carolina, 

7  per  cent. 

Georgia, 

8  per  cent. 

Alabama, 

8  per  cent. 

Mississippi, 

8  per  cent,  b 

Louisiana, 

5  per  cent,  c 

Tennessee, 

6  per  cent. 

Kentucky, 

6  per  cent. 

Ohio, 

6  per  cent. 

Indiana, 

6  per  cent. 

Illinois, 

6  per  cent,  d 

Missouri, 

6  per  cent,  e 

Michigan, 

7  per  cent. 

Arkansas, 

6  per  cent.  / 

Florida, 

8  per  cent. 

Wisconsin, 

7  per  cent,  g 

Iowa, 

7  per  cent,  h 

Texas, 

10  per  cent. 

Dist.  Columbia, 

6  per  cent. 

a  On  tobacco  contracts  8  per  cent,  b  By  contract  as  high  as  10  per  cent,  c  Bank  inter- 
est G  per  cent. ;  conventional  as  high  as  10  per  cent,  d  By  agreement  as  high  as  12  pet 
cent,  e  By  agreement  as  high  as  10  per  cent.  /  By  agreement,  any  rate  not  exceeding  10 
per  cent,  g  By  contract  as  high  as  12  per  cent,  h  By  agreement  as  high  as  12  per  cent. 


ARIS.  402, 403  ]  INTEREST.  251 

4O3.  Ex.  l.-What  is  the  interest  of  $30  for  1  year,  at  6 
per  cent.  ? 

Analysis. — We  have  seen  that  6  per  cent,  is  r^g- ;  that  is,  $6 
for  $100,  6  cents  for  100  cents,  &c.  (Art.  386.)  Since  therefore 
the  interest  of  $1  (100  cents)  for  1  year  is  6  cents,  the  interest  )f 
$30  for  the  same  time  must  be  30  times  as  much;  and  $30X-06 
=$1.80.  Ans. 

Operation.  We  first  multiply  the  principal   by  the 

$30  Prin.  given  rate  per  cent,  expressed  in  decimals, 

.06  Rate.  as  in  percentage,  and  point  off  as  many  de- 


$1.80  Int.  1  yr.       cimals  in  the  product  as  there  are  decimal 
places  in  both  factors. 

Ex.  2.  What  is  the  interest  of  $140.25  for  1  year,  1  month,  and 
10  days,  at  7  per  cent.  ?     What  is  the  amount? 

Operation. 

$140.25  Prin.  1  month  is  ^  of  a  year;  there- 

.07  Rate.  fore  the  interest  for  1  month  is  -fa 

12)$9.8175  Int.  1  yr.  of  1  year's  interest.     10  days  are  % 

3)      8181    "     1  mo.  of  1  month,  consequently  the  interest 

2727    "     10  d.  for  10  days,  is  -J-  of  1  month's  inter- 


$10.9083  Interest.  est.     The  amount  is  found  by  add- 

$140.25      Prin.  added,     ing  the  principal   and   interest   to- 
$151.1583  Amount.  gether. 

Note.— I.  In  adding  the  principal  and  interest,  care  must  be  taken  to  add 
dollars  to  dollars,  cents  to  cents,  &c.  (Art.  374.) 

2.  When  the  rate  per  cent,  is  Jess  than  10,  a  cipher  must  always.be  prefixed 
to  the  figure  denoting  it.  (Art.  387.  Obs.  1.)  It  is  highly  important  that  the 
principal  and  the  rate  should  both  be  written  correctly,  in  order  to  prevent  mis- 
takes in  pointing  off  the  product. 

Ex.  3.  What  is  the  interest  of  $250.80  for  4  years,  at  5  per 
cent.  9  What  is  the  amount  ? 

Szlution. — $250.80X.05=$12.54,  the  interest  for  1  year. 
Now     $12.54X     4=$50.16,       "  "        4  years. 

And   $250.80+$50.16=$300.96,  the  amount  required. 


252  INTEREST.  [SECT.  X.I1. 

4O  J;.  From  the  foregoing  illustrations  and,  principles  we  de- 
duce the  following  general 

RULE   FOR   COMPUTING  INTEREST. 

I.  FOR  ONE  YEAR.     Multiply  the  principal  by  the  given  rate,  and 
frwi  tlie  product  point  off  as  many  figures  for  decimals,  as  tluro, 
are  decimal  places  in  both  factors.  (Art.  324.) 

II.  FOR  TWO  OR  MORE  YEARS.     Multiply  the  interest  of  1  yew 
"by  tlie  given  number  of  years. 

III.  FOR  MONTHS.      Take  such  a  fractional  part  of  1  year's  in- 
terest, as  is  denoted  by  the  (jiven  number  of  months. 

IV.  FOR  DAYS.      Take  such  a  fractional  part  of  one  month's  in- 
terest, as  is  denoted  by  the  (jiven  number  of  days. 

The  amount  is  found  by  adding  the  principal  and  interest  together. 

OBS.  1.  The  reason  of  this  rule  is  evident  from  the  consideration  that  the 
given  rate  per  cent,  per  annum  denotes  kundrcdths.  (Arts.  386.  39S.)  Now 
when  the  rate  is  6  per  cent.  \ve  multiply  by  .OH,  when  7  per  cent,  by  .07,  &c., 
and  point  off  two  figures  in  the  product ;  consequently  the  result  will  be  the 
same  as  to  multiply  by  -y^p  -j-fo,  &c. 

2.  In  calculating  interest,  a  month,  whether  it  contains  30  or  31  days,  or 
even  but  28  or  29,  as  in  the  case  of  February,  is  assumed  to  be  one  twelfth,  of  a 
year.    Therefore,  for  1  month  we  take  -fa  of  1  year's  interest;  for  2  months,  -£-; 
for  3  months,  \ ;  for  4  months,  -^ ;  for  G  months,  •£ ;  for  8  months,  -f ,  &c. 

Again,  30  days  are  commonly  considered  a  month;  consequently  the  interest 
for  1  day,  or  any  number  of  days  under  30,  is  so  many  thirLieUis  of  a  month's 
interest.  (Art.  303.  Obs.  2.)  Therefore,  for  1  day  we  take  -^  of  1  month's 
interest ;  for  2  days,  -fa ;  for  3  days,  -fa ;  for  5  days,  •£• ;  for  10  days,  -£-,  &c. 

This  practice  seems  to  have  been  originally  adopted  on  account  of  its1,  con- 
venience. Though  not  strictly  accurate,  it  is  sanctioned  by  general  usage. 

3.  Allowing  30  days  to  a  month,  and  12  months  to  a  year,  a  year  would  con- 
tain only  300  days,  which  in  point  of  fact  is  -3-^3-,  or  -fa  less  than  an  ordinary 
year.     H:nce, 

To  find  the  interest  for  any  number  of  days  with  entire  accuracy,  we  must 
take  so  many  3G5ths  of  1  year's  interest,  as  is  demoted  by  the  given  number 
of  days ;  or.  find  the  interest  for  the  days  as  above  from  this  subtract  -fa  of 

QI-EST. — 404.  How  is  interest  computed  for  a  year?  I1,  w  for  any  number  of  years  J 
How  far  months?  How  for  days?  How  find  the  amoun'  1  Obs.  In  reckoning  interest, 
what  part  of  a  year  is  a  month  considered  ?  How  many* days  are  commonly  considered  a 
Bumth  ?  Ls  this  practice  accurate  ? 


Allf.  404.]  INTEREST.  253 

itself,  and  the  remainder  will  be  the  exact  interest.     The  laws  of  New  York, 
and  several  other  states,  require  this  deduction  to  be  made. 

In  business,  when  the  mills  in  the  result  are  5,  or  o\er,  it  is  customary  to 
add  I  to  the  cents ;  if  under  5,  to  disregard  them. 

EXAMPLES. 

]  What  is  the  interest  of  $423  for  1  yr.,  at  7  per  cent.  ? 

2.  What  is  the  interest  of  "§240.31  for  3  yrs.,  at  6  per  cent  ? 

3.  What  is  the  interest  of  $403.67  for  2  yrs.,  at  £  per  cent  ? 

4.  What  is  the  interest  of  $040  for  1  yr.,  at  8  per  cent  ? 

5.  What  is  the  interest  of  $430.45  for  2  yrs..  at  7  per  cent.  ? 

6.  What  is  the  interest  of  $185.00  for  4  yrs.,  at  0  per  cent.  ? 

7.  What  is  the  interest  of  $804.80  for  5  yrs.,  at  4|  per  cent.  ? 

8.  What  is  the  interest  of  $703  for  4  months,  at  7  per  cent.? 

9.  What  is  the  interest  of  $940.20  for  0  mo.,  at  0  per  cent.  ? 

10.  What  is  the  interest  of  $243.10  for  5  mo.,  at  8  per  cent.  ? 

11.  What  is  the  interest  of  $195.82  for  7  mo.,  at  0  per  cent.  ? 

12.  What  is  the  interest  of  $425.35  for  9  mo.,  at  0  per  cent.  ? 

13.  At  7  per  cent.,  what  is  the  int.  of  $738  for  1  yr.  and  2  mo.  ? 

14.  At  0  per  cent.,  what  is  the  int.  of  $894  for  1  yr.  and  8  mo.  ? 

15.  At  7  per  cent.,  what  is  the  amount  of  $920  for  0  mo.  ? 
10.  At  7  per  cent.,  what  is  the  amt.  of  $048  for  2  mo.  15  d.  ? 

17.  At  0  per  cent.,  what  is  the  amt.  of  $1000  for  1  mo.  lid.? 

18.  At  5  per  cent.,  what  is  the  amt.  of  $1505.45  for  3  mo.  ? 

19.  At  0  per  cent.,  what  is  the  amt.  of  $872  for  4  mo.  ? 

20.  What  is  the  int.  of  $081  for  10  days,  at  0  per  cent.  ? 

21.  What  is  the  int.  of  $483.20  for  15  d.,  at  7  per  cent.  ? 

22.  What  is  the  int.  of  $509.40  for  20  d.,  at  0  per  cent.  ? 

23.  What  is  the  amt.  of  $95  for  1  yr.  and  0  mo.,  at  5  percent.  ? 

24.  What  is  the  amt.  of  $148  for  8  mo.  ]2  d.,  at  0  per  cent  ? 

25.  What  is  the  amt.  of  $700  for  30  d.,  at  7  per  cent.  ? 
20.  What  is  the  int.  of  $340  for  00  d.,  at  5i  per  cent,  ? 

27.  What  is  the  int.  of  $4(585  for  90  d.,  at  0^  per  cent.  ? 

28.  What  is  the  amt.  of  $3293  for  30  d.,  at  7  per  cent.  ? 

29.  What  is  the  amt.  of  $5205  for  15  d.,  at  (:  per  cent.  ? 

30.  What  is  the  int.  of  $8310  for  10  d.,  at  7  per  cent.  ? 

31.  What  is  the  int.  of  $50025  for  21  d.,  at  7  per  cent.  ? 
82.  What  is  ihe  amt.  of  $05256  for  4  mo.,  at  7  per  cent.  ? 


254  INTEREST.  [SECT.  XII 

SECOND  METHOD   OF   COMPUTING   INTEREST. 

405.  There  is  another  method  of  computing  interest,  which 
is  very  simple  and  convenient  in  its  application,  particularly  when 
the  interest  is  required  for  months  and  days,  at  6  per  cent. 

406.  We  have  seen  that  for  1  year,  the  interest  of  81  at  6 
per  cent,  is  6  cents.,  or  $.06  ;  (Art.  404  ;)  therefore, 

For  1  month,  the  interest  of  $1  is  -^          of  6  cents,  which  is  $.005 ; 

"     2  months,            "             «   is  ^,  or -f  of  6  cents,      "      "  001; 

"     3  months,            «             «    is  -j^-,  or  |  of  6  cents,      "       '  .015; 

"     4  months,            "             "   is  -^,  or -^  of  6  cents,      "      «  .02 1 

"     5  months,            «             «    is  -j\,          of  G  cents,      "      "  .025; 

"     6  months,            "             "    is  -$j,  or  \  of  6  cents,      "      "  .03; 

Hence,  The  interest  of  $1  for  1  month,  at  6  per  cent.,  is  5  mills; 
for  every  2  months,  it  is  1  cent ;  and  for  any  number  of  months, 
it  is  as  many  cents,  or  hundredths  of  a  dollar,  as  2  is  contained 
times  in  the  given  number  of  montJis. 

407.  Since  the  interest  of  $1  for  1  month  (30  days)  is  5  mills, 
or  $.005,  (Art.  406,) 

For  6  days  (£  of  30  days)  the  interest  of  SI  is  £  of  5  mills,  or  $.001 ; 

«  12  days  (|  of  30  days)  «  "  is  -f  of  5  mills,  or    .002; 

«  18  days  (f  of  30  days)  «  "  is  -f  of  5  mills,  or    .003 ; 

"    3  days  (-fL  of  30  days)  "  is  fj  of  5  mills,  or  .0005 ; 

That  is,  the  interest  of  $1  for  every  6  days,  is  1  mill,  or  $.001 ; 
and  for  any  number  of  days,  it  is  as  many  mills,  or  thousandth 
of  a  dollar,  as  6  is  contained  times  in  the  given  number  of  days. 

4O8«  Hence,  to  find  the  interest  of  $1  for  any  number  of 
days,  at  6  per  cent. 

Divide  the  given  number  of  days  by  6,  and  set  the  first  quotient 
figure  in  thousandths'  place,  when  the  days  are  6,  or  more  than  6 ; 
but  in  ten  thousandths'  place,  when  they  are  less  than  6. 

OBS.  For  60  days  (2  mo.)  the  interest  of  $1  is  1  cent;  (Art.  406,;  v*  nen, 
therefore,  the  number  of  days  is  60  or  over,  the  first  quotient  figure  must 
occupy  hundredth^  place. 

QUEST. — 408.  How  find  the  interest  of  $1  for  any  number  of  days,  at  6  per  cent.  1 


ARTS.  405 -409. J  INTEREST.  255 

Ex.  1.  What  is  the  interest  of  $185  for  1  year,  6  months  and 
18  days,  at  6  per  cent.? 

Analysis. — The  interest  of  $1  for  1  year  is  ,    Operation. 
6  cents  ;  for  6  months  it  is  3  cents;  and  for  $185  Prill. 

18  days  it  is  3  mills.   (Arts.  406,  407.)     Now  .093  Int.$l. 

.06  + .03+.003=$.093.     Since  therefore  the  555 

interest  of  $1  for  the  given  time  is  $.093,  the          1665 

interest  of  $185  must  be  185  times  as  much.  $17.205  Ans. 

4O9«  Fr<  m  these  principles  we  may  derive  a 

SECOND  RULE  FOR  COMPUTING  INTEREST. 

1.  To  compute  the  interest  on  any  sum,  at  6  per  cent. 
Multiply  the  principal  by  the  interest  of  $1  for  the  given  time, 

at  6  per  cent.,  and  point  off  the  product  as  in  multiplication  of 
decimals.  (Art.  324.) 

II. .  To  compute  int.  at  any  rate,  greater  or  less  than  6  per  cent. 

First  find  the  interest  on  the  given  sum  at  6  per  cent,  j  then, 
add  to  this  interest,  or  subtract  from  it,  such  a  fractional  part  of 
itself,  as  the  required  rate  exceeds  or  falls  short  of  6  per  cent. 

The  amount  is  found  by  adding  the  principal  and  interest  to- 
gether as  in  the  former  method.  (Art.  404.) 

Ops.  ' .  The  amount  may  also  be  found  by  multiplying  the  given  principal  by 
the  amount  of  one  dollar  for  the  time. 

2.  The.  rental  of  the  first  part  of  this  rule,  is  manifest  from  the  principle  that 
ihe  interest  of  2  dollars  for  any  given  time  and  rate,  must  be  twice  as  much  as 
the  interest  of  1  dollar  for  the  same  time  arid  rate ;  the  interest  of  50  dollars, 
50  times  as  much  as  that  of  1  dollar,  &c. 

3.  When  the  required  rate  is  7  per  cent.,  we  first  find  the  interest  at  6  per  cent., 
then  add  -£•  of  it  to  itself;  if  5  per  cent.,  subtract  4-  of  it  from  itself,  &c.,  for  the 
obvious  reason,  that  7  per  cent,  is  once  and  1  sixth,  or  -^  of  G  per  cent. ;  5 
per  cent,  is  only  A  of  6  per  cent.,  &c. 

\.  When  the  decimal  denoting  the  int.  of  $'1  for  the  days,  is  long,  or  is  a  repa- 
tend,  it  is  more  accurate  to  retain  the  common  fraction.  (Art.  387.  Obs.  2.) 

2.  What  is  the  interest  of  $746  for  4  months  and  18  days,  at 
6  pei  ont.?  Ans.  $17.153. 

Quasi. — 409.  What  is  the  second  method  of  computing  interest,  at  6  per  cent.  ?  Whea 
the  ti.te  uei  cent,  is  greater  or  less  than  6  per  cent.,  how  proceed  ? 


2,56  INTEREST.  [SECT.  XII. 

3.   What  is  the  interest  of  $240  for  G  months  and  12  days,  at 
7  pei  cent.  ? 

Operation. 

£240  Pnn.  The  interest  of  $1  for  6  mo.  at  6 

.032  Int.  of  $1.  per  ct.,  is  .03  ;  for  12  d.  it  is  .002 ; 

480                           .  and  .03  +  .002=8.032. 

720  The  required  rate  is  1  per  cent. 

6)$7.680=Int.  at  6  perct.  more  than  6  per  cent. ;  we  there- 

1.280=£  of  6  per  ct.  fore  find  the  interest  at  6  per  cent., 

Ans.  $8.960  Int.  at  7  per  ct.  and  add  |  of  it  to  itself. 

4.  What  is  the  interest  of  $t>80  for  3  mo.,  at  5  per  cent.  ? 

5.  What  is  the  interest  of  $213.08  ibr  1  mo.,  at  6  per  cent.  ? 

6.  What  is  the  interest  of  $859  for  1  yr.  2  mo.,  at  7  per  cent.*? 

7.  What  is  the  interest  of  $708  for  1  yr.  7  mo.,  at  8  per  cent.? 

8.  What  is  the  interest  of  $684  for  9  mo.,  at  6  per  cent.  ? 

9.  At  7  per  cent.,  what  is  the  amount  of  $387  for  5  mo.  ? 

10.  At  4  per  cent.,  what  is  the  amt.  of  $1125  for  1  yr.  2  mo.  ? 

11.  At  6  per  cent.,  what  is  the  amt.  of  $1056  for  10  mo.  24  d.  ? 

12.  At  6  per  cent.,  what  is  the  int.  of  $1340  for  1  mo.  15  d.  ? 

13.  At  6  per  cent.,  what  is  the  int.  of  $815  for  2  mo.  21  d.  ? 

14.  At  8  per  cent.,  what  is  the  amt.  of  $961  for  4  mo.  10  d.  ? 

15.  What  is  the  int.  of  $2345.10  for  6  mo.,  at   7   per  cent.  ? 

16.  What  is  the  int.  of  $1567.18  for  4  mo.,  at  7£  per  cent.  ? 

17.  What  is  the  int.  of  $3500  for  11  mo.,  at  10  per  cent.  ? 

18.  What  is  the  int.  of  $39.375  for  2  yrs.,  at  12£  per  cent.  ? 

19.  What  is  the  int.  of  $113.61  for  5  yrs.,  at  15  per  cent.? 

20.  What  is  the  int.  of  $1000  for  2  yrs.,  at  20  per  cent.  ? 

21.  What  is  the  int.  of  $1260.34  for  10  yrs.,  at  13  per  cent.  ? 

22.  At  16  per  cent,,  what  is  the  int.  of  $150  for  6  years.  ? 

23.  At  30  per  cent.,  what  is  the  int.  of  $300  for  1  year.  ? 

24.  What  is  the  amt.  of  $12645  for  10  i,  at  6  per  cent.  ? 

25.  What  is  the  amt.  of  $16285  for  24  1,  at  7  per  cent.  ? 

26.  At  4^  per  cent.,  what  is  the  int.  of  $10255  for  8  months  ? 

27.  At  5|  per  cent.,  what  is  the  int.  of  $17371  for  3  months  ? 

28.  What  is  the  amt.  of  $1  for  100  yrs.,  at  7  per  cent.  ? 

29.  What  is  the  amt.  of  1  cent  for  100  yrs.,  at  6  per  cent.  ? 


ARTS.  410,  411. J  INTEREST.  257 

41 0.  Since  the  interest  of  $1  at  G  per  cent,  for  12  rno.  is  6 
cents,  (Art.  406,)  for  G  mo.  it  must  be  3  cents ;  for  3  mo.,  !-£  cents ; 
for  2  mo.,  1  cent;  for  1  mo.  or  30  d.  \  cent;  for  15  d.,  |  cent; 
for  20  d.  i-  cent,  &c.     That  is,  the  interest  of  $1  at  G  per  cent. 
is  as  many  cents  as  are  equal  to  half  the  given  number  of  months. 

411.  Hence,  to  compute  interest  at  6  per  cent,  by  montlis. 
Multiply  the  principal  by  half  the  number  of  months,  and  point 

'}Jf'  two  more  figures  for  decimals  in  tJie  product  than  there  are  deci- 
mal places  in  the  multiplicand. 

OBS.  1.  When  there  are  years  and  days,  reduce  the  years  to  months,  and 
the  days  to  a  common  fraction  of  a  month. 

Or,  divide  the  days  by  3,  and  annex  the  quotient  to  the  months  considered 
as  hundredth*;  half  of  the  number  thus  produced  will  be  the  decimal  multiplier 

2.  The  latter  method  is  the  same  as  dividing  the  days  by  6,  and  setting  the 
fii  st  quotient  figure  in  thousandth's  place ;  for,  we  divide  the  days  by  3  and 
2,  and  3X2=6.  (Arts.  407,  408.) 

30.  What  is  the  int.  of  $460.384  for  8  mos.  and  15  d.,  at  6  per  ct.  ? 

Operation. 

$460.384  We  multiply  by  4|,  for,  8  months-}- 15 

4f  days=8i  months,  and  8^-f-2  =  4|.     And 

1841536  since  there  are  three  decimals  in  the  mul- 

115096  tiplicand,  we  point  off  5  in  the  product. 
$19.56632  Am. 

31.  What  is  the  interest  of  $780  for  4  months,  at  6  per  cent.  ? 

32.  What  is  the  interest  of  $1406  for  3  mo.,  at  6  per  cent.  ? 

33.  What  is  the  interest  of  $109  for  2  mo.,  at  7  per  cent.  ? 

34.  What  is  the  interest  of  $119.45  for  8  mo.,  at  6  per  cent.  ? 

35.  What  is  the  interest  of  $618  for  1  yr.  3  mo.,  at  6  percent.  ? 

36.  Wha',  is  the  interest  of  $861  for  2  yrs.  6  mo.,  at  6  per  cent.  ? 

37.  What  is  the  interest  of  $936.40  for  3  yrs.,  at  6  per  cent.  ? 

38.  What  is  the  interest  of  $4526  for  6  mo.  2  d.,  at  6  per  cent.  ? 

39.  What  is  the  interest  of  $8246  for  10  mo.,  at  7  per  cent.  ? 

40.  What  is  the  interest  of  $31285  for  3  mo.,  at  5  per  cent.  ? 

41.  What  is  the  interest  of  $17500  for  1  yr.  3  mo.,  at  7  per  ct.  ? 

42.  What  is  the  amount  of  $3286  for  8  mo.  15  d.,  at  6  per  ct.  t 

43.  What  is  the  amount  of  $15876  for  5  mo.  18  d.,  at  6  per  ct.  ? 


258  APPLICATIONS    OF  [SECT.   XIl. 

412.  We  have  seen  that  the  interest  of  $1  at  6  per  cent,  for 
any  number  of  days  is  equal  to  as  many  mills,  as  6  is  contained 
times  in  the  given  days.  (Art.  407.)     Hence, 

413.  To  compute  interest  at  6  per  cent,  by  days. 
Multiply  the  principal  by  one  sixth  of  the  given  number  of  days, 

and  point  off  three  more  figures  for  decimals  in  the  product  t/tan 
tlicre  are  decimal  places  in  the  principal.  (Art.  411.  Obs.  2.) 

Or,  mdltiply  the  principal  by  the  given  number  of  days,  divid 
the  product  by  6,  and  point  off  the  quotient  as  above. 

OBS.  The  product  is  in  mills  and  parts  of  a  mill.  The  object,  therefore,  of 
pointing  off  three  more  places  for  decimals  in  the  product  than  there  are  deci- 
mals in. the  principal,  is  to  reduce  it  to  dollars.  (Art.  372.) 

44.  What  is  the  interest  of  1976.22  for  33  days,  at  6  per  cent.  ? 
Solution. — |  of  33  d.  =  5i-;  and  $976.22X5-^=5369.21  mills. 

Pointing  off  3  more  decimals,  we  have  $5.36921.  Ans. 

45.  What  is  the  interest  of  $536.30  for  24  days,  at  6  per  cent.  ? 

46.  What  is  the  interest  of  $7085  for  63  d.,  at  6  per  cent.  ? 

47.  What  is  the  interest  of  $8126.21  for  8  d.,  at  6  per  cent.  ? 

48.  What  is  the  interest  of  $25681  for  93  d.,  at  6  per  cent.  ? 

49.  What  is  the  interest  of  $764.85  for  114  d.,  at  6  per  cent.  ? 

APPLICATIONS    OP   INTEREST. 

414.  In  the  application  of  interest  to  business  transactions, 
the  following  particulars  deserve  attention. 

1.  A  promissory  note  is  a  writing  which  contains  a  promise  of  the  payment 
of  money  or  other  property  to  another,  at  or  before  a  time  specified,  in  consid- 
eration of  value  received  by  the  promiser  or  maker  of  the  note. 

Unless  a  note  contains  the  words  "  value  received,"  by  some  authorities  it 
is  deemed  invalid ;  consequently  these  words  should  always  be  inserted. 

2.  The  person  who  signs  a  note  is  called  the  r.iaker,  drawer,  or  giver  of  the 
note.     The  person  to  whom  a  note  is  made  payable,  is  called  the  payee;  the 
person  who  has  the  legal  possession  of  a  note,  is  called  the  Jwlder  of  it. 

3  A  note  which  is  made  payable  "  to  order ',"  "  or  bearer'*  is  said  to  be  ncgo 
tiab.'e ;  that  is,  the  holder  may  sell  or  transfer  it  to  whom  he  pleases,  and  it  can 
be  collected  by  anyone  who  has  lawful  possession  of  it.  Notes  without  these 
words  are  not  negotiable.  (See  Nos.  1,  2.) 

4.  If  the  holder  of  a  negotiable  note  which  is  made  payable  to  order  wishes 
to  sell  or  transfer  it,  the  law  requires  him  to  endorse  it,  or  write  his  name  on 
the  back  of  it.  The  person  to  whom  it  is  transferred,  or-the  holder  of  it,  ia 


A.UTS.   4  J  2-4 15.]  INTEREST.  259 

then  empowered  to  collect  it  of  the  drawer;  if  the  drawer  is  unable,  or  icfuset 
to  pay  it,  then  the  endorser  is  responsible  for  its  payment.  (See  No.  1.) 

5.  When  a  note  is  made  payable  to  the  bearer,  the  holder  can  sell  or  trans- 
fer it  without  endorsing  it,  or  incurring  the  liability  lor  its  payment.     Bank 
notes  or  bills  are  of  this  description.  (See  No.  2.) 

6.  When  a  note  is  made  payable  to  any  particular  person  without  the  words 
order  or  bearer  it  is  nob  negotiable;  for,  it  cannot  be,  collected  or  sued  except  in 
the  name^j^he  person  to  whom  it  is  made  payable.  (See  No.  3.) 

7.  A  note  should  always  specify  the  time  at  which  it  is  to  be  paid;  but  if  no 
time  is  mentioned,  the  presumption  is  that  it  is  intended  to  be  paid  on  demand, 
and  the  giver  must  pay  it  when  demanded. 

8.  According  to  custom  and  the  statutes  of  most  of  the  States,  a  note  or 
draft  is  not  presented  for  collection  until  ikrcc  days  after  the  time  specified  for 
its  payment.     These  three  days  are  called  dai/s  of  grace.     Interest  is  therefore 
reckoned  for  three  days  more  than  the  time  specified  in  the  note.     When  the 
last  day  of  grace  comes  on  Sunday,  or  a  national  holiday,  as  the  4th  of  July, 
&c..  it  is  customary  to  pay  a  note  on  the  day  previous. 

9.  If  a  note  is  not  paid  at  maturity  or  the  time  specified,  it  is  necessary  for 
the  holder  to  notify  the  endorser  of  the  fact  in  a  legal  manner,  as  soon  as  cir- 
cumstances will  admit ;  otherwise  the  responsibility  of  the  endorser  ceases. 

10.  Notes  do  not  draw  interest  unless  they  contain  the  words  "  with  inter- 
est."    But  if  a  note  is  not  paid  when  it  becomes  due,  it  then  draws  legal  in- 
cerest  till  paid,  though  no  mention  is  made  of  interest.  (Art,  400.  Obs.) 

11.  Notes  which  contain  the  words  "  with  interest"  though  the  rate  is  not 
mentioned,  are  entitled  to  the  legal  rate  established  by  the  State  in  which  the 
note  is  made.     In  writing  notes  therefore  it  is  unnecessary  to  specify  the  rate, 
unless  by  agreement  it  is  to  be  less  than  the  legal  rate. 

12.  When  a  note  is  made  payable  on  a  given  day,  and  in  a  specified  article 
of  merchandise,  as  grain,  stock,  &c.,  if  the  article  specified  is  not  tendered  at 
the  given  time  and  place,  the  holder  can  demand  payment  in  money.     Such 
notes,  are  not  negotiable ;  nor  is  the  drawer  entitled  to  the  days  of  grace. 

13.  When  two  or  more  persons  jointly  and  severally  give  their  note,  it  may 
be  collected  of  either  of  them.  (See  No.  4.) 

14.  The  sum  for  which  a  note  is  given,  is  called  the  principal,  or  face  oft\it 
note;  and  should  always  be  written  out  in  words. 

415*  When  it  is  required  to  compute  the  interest  on  a  note, 
we  must  first  find  the  time  for  which  the  note  has  been  on  inter 
est,  by  subtracting  the  earlier  from  the  later  date;  (Art.  303; 
then  cast  the  interest  on  the  face  of  the  note  for  the  time,  by 
either  of  the  preceding  methods.  (Arts.  404,  409.) 

OBS.  In  determining  the  time,  the  day  on  which  a  note  is  dated,  and  tlu't 
on  which  it  becomes  due  should  not  both  be  reckoned ;  it  is  customary  to  ex« 
elude  the  former. 

T.H.  12 


260  APPLICATIONS    OF  [SECT.   XII. 

Ex.  1.  What  is  the  interest  due  on  a  note  j>f  $625  frcm  FeL  2d, 
1846,  to  June  20th,  1847,  at  6  per  cent.? 

Ojxration.  $625  Prin. 


Yrs.          mo.        ds. 

1847  "  6  "  20 

1846"2"     2 
1   "  4  "  18 


Int  of  $1. 


5000 


$5L875  Ans. 
Compute  the  interest  on  the  following  notes  : 

.  _  (No.  1.) 

$450.  NEW  YORK,  June  3d,  1847. 

2.  Sixty  days  after  date,  I  promise  to  pay  George  Baker,  or 
order,  Four  Hundred  and  Fifty  Dollars,  with  interest,  value  re- 
ceived. ALEXANDER  HAMILTON. 

_  (No.  2.) 

$630.  BOSTON,  Aug.  5th,  1847. 

.  3.  Thirty  days  after  date,  I  promise  to  pay  Messrs.  Holmes  & 
Homer,  or  bearer,  Six  Hundred  and  Thirty  Dollars,  with  interest, 
value  received.  JAMES  UNDERWOOD. 

(No.  3.) 


$850.  PHILADELPHIA,  Sept.  16th,  1847. 

4.  Four  months  after  date,  I  promise  to  pay  Horace  Williams, 
Eight  Hundred  and  Fifty  Dollars,  with  interest,  value  received. 

JOHN  C.  ALLEN. 

(No.  4.) 
$1000.  CINCINNATI,  Oct.  3d,  1847. 

5.  For  value  received,  we  jointly  and  severally  promise  to  pay 
to  the  order  of  Wm.  D.  Moore  &  Co.,  One  Thousand  Dollars,  in 
one  year  from  date,  with  interest.  JOSEPH  HENRY, 

SANDFORD  ATWATER. 

6.  What  is  the  interest  on  a  note  of  $634  from  Jan.  1st,  1 846, 
to  March  7th,  1847,  at  6  per  cent.  ? 


ARTS.  415,  416.]  INTEREST.  201 

7.  What  is  the  interest  on  a  note  of  $820  from  April  16th, 
1846,  to  Jan.  10th,  1847,  at  6  per  cent.  ? 

8.  What  is  the  interest  on  a  note  of  $615.44  from  Oct.  1st, 
1836,  to  June  13th,  1840,  at  4  per  cent.  ? 

9.  What  is  the  interest  on  a  note  of  $1830.63  from  Aug.  16th, 
1841,  to  June  19th,  1842,  at  7  per  cent.  ? 

10.  What  is  the  amount  due  on  a  note  of  $520  from  Sept.  2d, 
1846,  to  March  14th,  1847,  at  5  per  cent.  ? 

1 1.  What  is  the  amount  due  on  a  note  of  $25000  from  Aug. 
17th,  1845,  to  Jan.  17th,  1846,  at  7  percent? 

12.  What  is  the  amount  due  on  a  note  of  $6200  from  Feb.  3d, 
1846.  to  Jan.  9th,  1847,  at  6  per  cent.? 

PARTIAL   PAYMENTS. 

416»  When  partial  payments  are  made  and  endorsed  upon 
Notes  and  Bonds,  the  rule  for  computing  the  interest  adopted  by 
the  Supreme  Court  of  the  United  States,  is  the  following. 

I.  "  The  rule  for  casting  interest,  when  partial  payments  have 
been  made,  is  to  apply  the  payment,  in  tJie  first  place,  to  the  dis- 
charge of  the  interest  then  due. 

II..  "  If  the  payment  exceeds  the  interest,  the  surplus  goes  towards 
discharging  the  principal,  and  the  subsequent  interest  is  to  be  com- 
puted on  the  balance  of  principal  remaining  due. 

III.  "  If  the  payment  be  less  than  the  interest,  the  surplus  of 
interest  must  not  be  taken  to  augment  the  principal  /  but  interest 
continues  on  the  former  principal  until  the  period  when  the  pay- 
ments, taken  together,  exceed  tlie  interest  due,  and  then  the  surplus 
is  to  be  applied  towards  discharging  tlie  principal ;  and  interest  is 
to  be  computed  on  the  balance  as  aforesaid" 

Note. — The  above  rule  is  adopted  by  New  York,  MassachiLsetts,  and  most 
if  the  other  States  of  the  Union.  It  is  given  in  the  language  of  the  distin- 
grdshed  Chancellor  Kent.— Johnson's  Chancery  Reports,  Vol.  I.  p.  17. 

UUHST. — 416.  What  is  the  genera\  method  of  casting  interest  on  Notes  and  Bonds,  Then 
partial  pay  merits  have  been  made  ? 


262  APPLICATIONS    OF  [SECT.    XII 

$965.  NEW  YORK,  March  8th,  1843.. 

13  For  value  received,  I  promise  to  pay  George  B.  Granniss, 
or  order,  Nine  Hundred  and  Sixty-five  Dollars,  on  demand,  with 
interest  at  7  per  cent.  HENRY  BROWN. 

The  following  payments  were  endorsed  on  this  note : 

Sept.  8th,  1843,  received  $75.30. 

June  18th,  1844,  received  $20.38. 

March  24th,  1845,  received  $80. 
What  was  due  on  taking  up  the  note,  Feb.  9th,  1846  ? 

Operation. 

Principal,                                                                 -  1*965.00 

Interest  to  first  payment,  Sept.  8th.  (6  months,)  33.775 

Amount  due  on  note  Sept.  8th,  $998.775 

1st  payment,  (to  be  deducted  from  amount,)       -  75.30 

Balance  due  after  1st  pay't.,  Sept.  8th,  1843,       -  $923.475 
Interest  on  Balance  to  2d  pay't.,  June  )      ^Kf.  n*0 

18th,  (9  mo.  10  d.,)  1 

2d  pay't.,  (being  less  than  int.  then  due,)          20.38 
Surplus  int.  unpaid  June  18th,  1844,  $29.898 

Int.  continued  on  Bal.  from  June  18th,  ) 

to  March  24th,  1845,  (9  mo.  6  d.,)     $        49'559 79'457 

Amount  due  March  24th,  1845,                            -  $1002.932 
3d  pay't.,  (being  greater  than  the  int.  now  due,)  > 

is  to  be  deducted  from  the  amount,  >         ' 

Balance  due  March  24th,  1845,  $922.932 

.     Int.  on  Bal.  to  Feb.  9th,  (10  mo.  15  d.,)     -  56.529 

Bal.  due  on  taking  up  the  note,  Feb.  9th,  1846,  $979.461 

$650.  BOSTON,  Jan.  1st,  1842. 

14.  1  or  value  received,  I  promise  to  pay  John  Lincoln,  or 
orler,  Six  Hundred  and  Fifty  Dollars  on  demand,  with  interest 
at  8  per  cent.  GEORGE  LEWIS. 

Endorsed,  Aug.  13th,  1842,  $100. 
Endorsed,  April  13th,  1843,  $120. 
What  was  due  on  the  note,  Jan.  20th,  1844? 


ART.  417.]  INTEREST.  263 

$2460,  PHILADELPHIA,  April  10th,  1844. 

45.  Four  months  after  date,  I  promise  to  pay  James  Buchanan, 
or  order,  Two  Thousand  Four  Hundred  and  Sixty  Dollars,  with  in- 
terest, at  6  per  cent.,  value  received. 

GEORGE  WILLIAMS. 

Endorsed,  Aug.  20th,  1845,  $840. 
Dec.  26th,  1845,    $400. 
May  2d,  1846,     $1000. 
How  much  was  due  Aug.  20th,  1846  ? 

$500a  NEW  ORLEANS,  May  1st,  1845. 

16.  Six  months  after  date,  I  promise  to  pay  John  Fairfield,  or 
order,  Five  Thousand  Dollars,  with  interest  at  5  per  cent.,  value 
received.  WILLIAM  ADAMS. 

Endorsed,  Oct.  1st,  1845,       $700. 
"          Feb.  7th,  1846,        $45. 
Sept.  13th,  1846,  $480. 
What  was  the  balance  due  Jan.  1st,  1S47  ? 

CONNECTICUT  RULE. 

417.  I.  "  Compute  the  interest  on  the  principal  to  the  time  of  the  first  pay- 
ment ;  if  that  be  one  year  or  more  from  the  time  the  interest  commenced,  add 
it  to  the  principal,  and  deduct  the  payment  from  the  sum  total.  If  there  be 
after  payments  made,  compute  the  interest  on  the  balance  due  to  the  next  pay- 
ment, and  then  deduct  the  payment  as  above ;  and  in  like  manner,  from  one 
payment  to  another,  till  all  the  payments  are  absorbed;  provided  the  time  be- 
tween one  payment  and  another  be  one  year  or  more." 

II.  "  If  any  payments  be  made  before  one  year's  interest  has  accrued,  then 
compute  the  interest  on  the  principal  sum  due  on  the  obligation,  for  one  year, 
add  it  to  the  principal,  and  compute  the  interest  on  the  sum  paid,  from  the 
time  it  was  paid  up  to  the  end  of  the  year ;  add  it  to  the  sum  paid,  and  deduct 
that  sum  from  the  principal  and  interest  added  as  above." 

III.  "  If  a  year  extends  beyond  the  time  of  payment,  then  find  the  tun  tint 
of  the  principal  remaining  unpaid  up  to  the  time  of  settlement,  likewise  the 
amount  of  the  endorsements  from  the  time  they  were  paid  to  the  time  of  settle- 
ment, and  deduct  the  sum  of  these  several  amounts  from  the  amount  of  the 
principal." 

"  If  any  payments  be  made  of  a  'ess  sum  than  the  interest  arisen  at  the  time 
of  such  payment,  no  interest  is  to  e  computed,  but  only  on  the  principal  sum 
for  any  period."— Kirby's  Reports. 


264  APPLICATIONS    OF  [SECT.  XIL 

THIRD  RULE. 

4 1  §.  First  find  the  amount  of  the  given  principal  for  the  whole  time ;  then 
find  live  amount  of  each  payment  from  the  time  it  was  endorsed  to  ttvc  time  of 
settlement.  Finally,  subtract  the  amount  of  the  several  payments  from  iht 
amount  of  the  principal,  and  the  remainder  will  be  Uic  sum  due. 

Note.—\i  will  be  an  excellent  exercise  for  the  pupil  to  cast  the  interest  on 
the  preceding  notes  by  each  of  the  above  rules. 

419.   To  compute  Interest  on  Sterling  Money. 

17.  What  is  the  interest  of  £241,  10s.  Gd.  for  1  year,  at  6  pei 
cent.  ? 

Operation. 

£241.525  Prin.  We  first  reduce  the  10s.  6d.  to  the 

.06  Rate.  decimal  of  a  pound,  (Art.  346,)  then 

£14.49150  Int.  1  yr.  multiply  the   principal  by  the  rate, 

20s.=£l.  and  point  off  the  product  as  in  Art. 

s.  9.83000  404.     The  14  on  the  left  of  the  deci- 

12d.=ls.  mal  point,  denotes  pounds;  the  fig- 

d.  9.96000  ures  on  the  right  are  decimals  of  a 

4f.=ld.  pound,  and  must  be  reduced  to  shil- 

far.  3.84000  lings,  pence,  and  farthings.  (Art.  348.) 

Ans.  £14,  9s.  9fd.  Hence, 

419.  a.  To  compute  the  interest  on  pounds,  shillings,  pence, 
and  farthings. 

Reduce  the  given  shillings,  pence,  and  farthings  to  the  decimal 
of  a  pound  ;  (Art.  346  ;)  then  find  tlie  interest  as  on  dollars  and 
cents  ;  finally,  reduce  the  decimal  figures  in  the  answer  to  shillings, 
pence,  and  farthings.  (Art.  348.) 

18.  What  is  the  amount  of  £156,  15s.  for  1  year  and  4  months, 
at  5  per  cent.  ?  Ans.  £167,  4s. 

19.  What  is  the  int.  of  £275,  12s.  6d.  for  1  yr.,  at  7  per  cent. 

20.  What  is  the  int.  of  £89,  7s.  6id.  for  2  yrs.,  at  5  percent. 

21.  What  is  the  int.  of  £500  for  6  mo.,  at  5  per  cent.  ? 

22.  What  is  the  amt.  of  £1825,  10s.  for  8  mo.,  at  6  per  cent.  ? 

23.  What  is  the  amt.  of  £2000  for  10  yrs.,  at  4|  per  cent.  ? 

QUEST. — 419.  How  is  interest  computed  on  pounds,  shillings,  and  peace  1 


ARTS.  418-421. J  INTEREST.  265 

PROBLEMS   IN  INTEREST. 

4  2O.  It  will  be  observed  that  there  are  four  parts  or  terms 
connected  with  each  of  the  preceding  operations,  viz :  t/ie  prind~ 
pal,  the  rate  per  cent.,  the  time,  and  the  interest,  or  the  amount. 
These  parts  or  terms  have  such  a  relation  to  each  other,  that  if 
any  three  of  them  are  given,  the  other  may  be  found.  The  ques- 
tions, therefore,  which  may  arise  in  interest^  are  numerous ;  but 
tiny  may  be  "reduced  to  a  few  general  principles,  or  Problems. 

OE?  A  number  or  quantity  is  said  to  be  given,  when  its  value  is  stated;  or 
may  bo  ea»ily  inferred  from  the  Conditions  of  the  question  under  consideration. 
Thus,  when  the  principal  and  interest  are  known,  the  amount  may  be  said  to 
be  given,  because  it  is  merely  the  sum  of  the  principal  and  interest.  So,  if  the 
principal  ind  the  amount  are  known,  the  interest,  may  be  said  to  be  given,  be- 
cause it  is  the  difference  between  the  amount  and  the  principal. 

PROBLEM    I. 

421.  To  find  tJte  INTEREST,  the  principal,  rate  per  cent.,  and 
the  time  being  given. 

This  problem  embraces  all  the  preceding  examples  pertaining 
to  Interest,  and  has  already  been  illustrated. 

PROBLEM    II. 

To  find  the  RATE  PER  CENT.,  t/ie  principal,  the  interest,  and  the 
time  being  given. 

Ex.  1.  A  man  borrowed  $80  for  5  years,  and  paid  $3G  for  the 
use  of  it :  what  was  the  rate  per  cent.  ? 

Analysis. — The  interest  of  $80  at  1  per  cent,  for  1  year  is  80 
cents  ;  (Art.  404  ;)  consequently  for  5  years  it  is  5  times  as  much, 
and  $.80  X  5— $4.  Now  since  $4  is  1  per  cent,  on  the  principal  for 
the  given  time,  $3G  must  be  ^  of  1  per  cent.,  which  is  equal  to 
9  per  cent.  (Art.  190.) 

Or,  we  may  reason  thus :  Since  $4  is  1  per  cent,  on  the  princi- 
pal for  the  given  time,  $36  must  be  as  many  per  cent,  as  $4  is 
contained  times  in  $36  ;  and  $36-r-$4=9.  Ans.  9  per  cent. 

QUKST — 420.  How  many  terms  are  connected  with  each  of  the  preceding  examples  1 
What  are  they  1  When  three  are  given,  can  the  fourth  be  found  ?  Obs.  When  is  a  nun* 
her  or  auantity  said  to  be  given  ? 


266  APPLICATIONS    OF  [SECT.  XIL 

PROOF $80X-09=$7.20,  the  interest  of  $80  for  1  year  at 

9  per  cent.,  and  $7.20X5=$36.00,  the  interest  for  5  years,  which 
is  equal  to  the  sum  paid.     Hence, 

422*  To  find  the  rate  per  cent,  when  the  principal,  interest,' 
and  time  are  given. 

Divide  the  given  interest  by  the  interest  of  the  principal  at  1 
per  cent,  for  tJie  given  time,  and  the  quotient  will  be  the  required 
per  cent. 

Or,  find  the  interest  of  the  principal  at  1  per  cent  for  the 
given  time  ;  then  make  the  interest  thus  found  the  denominator  and 
the  given  interest  tlie  numerator  of  a  common  fraction  ;  reduce  this 
fraction  to  a  whole  or  mixed  number,  and  the  result  wilt  be  the  per 
cent,  required.  (Art.  196.) 

2.  If  I  loan  $500  for  2  years,  and  receive  $50  interest,  what  is 
the  rate  per  cent.  ?  Ans.  5  per  cent. 

3.  A  man  borrowed  $620  for  8  months,  and  paid  $24.80  for 
the  use  of  it :  what  per  cent,  interest  did  he  pay  ? 

4.  At  what  per  cent,  interest  must  $2350  be  loaned,  to  gain 
$£7  in  4  months  ? 

'5.  At  what  per  cent,  interest  must  $1925  be  loaned,  to  gain 
$154  in  1  year? 

6.  A  man  has  $12000  from  which  he  receives  $900  interest 
annually :  what  per  cent,  is  that  ? 

7.  A  man  deposited  $2600  in  a  savings  bank,  and  received  $143 
interest  annually :  what  per  cent,  was  that  ? 

8.  A  man  invested  $4500  in  the  Bank  of  New  York,  and  re- 
ceived a  semi-annual  dividend  of  $lf>7.50  :  what  per  cent,  was  tht 
dividend  ? 

9.  A  man  paid  $16250  for  a  house,  and  rented  it  for  $975  a 
year :  what  per  cent,  did  it  pay  ? 

10.  A  hotel  which  cost  $250000,  was  rented  for  $125'X)  a  year: 
•what  per  cent,  did  it  pay  on  the  cost? 

11.  A  capitalist  invested  $500000  in  manufacturing,  and  re- 
ceived a  semi-annual  dividend  of  $12500  :  what  per  cent,  was  his 
dividend  ? 

ttoisT.— 422.  When  the  principal,  interest  and  time  are  given,  how  is  flie  rite  per  ct.  found  f 


ARTS.  422,  423.]  INTEREST.  267 

PROBLEM    III. 

To  find  the  PRINCIPAL,  t/ie  interest,  the  rate  per  cent.,  and  the 
time  being  given. 

« 

12.  What  sum  must  be  put  at  interest,  at  6  per  cent.,  to  gam 
$75  in  2  years  ? 

Analysis. — The  interest  of  $1  for  2  years  at  6  per  cent.,  (the 
g  ~en  time  and  rate,)  is  12  cents.  Now  12  cents  interest  is  -^fc 
ot  its  principal  $1 ;  consequently,  $75,  the  given  interest,  must  be 
•ffi;  of  the  principal  required.  The  question  therefore  resolves 
itself  into  this :  $75  is  iVo"  of  what  number  of  dollars  ?  If  £75  is 
iW,  T-Jr  is  A  of  175,  which  is  $6|;  and  +££=461x100,  which 
is  $625,  the  principal  required. 

Or,  we  may  reason  thus :    Since  1 2  cents  is  the  interest  of  1 
dollar  for  the  given  time  and  rate,  75  dollars  must  be  the  interest 
of  as  many  dollars  for  the  same  time  and  rate,  as  12  cents  is  con 
tained  times  in  75  dollars.     And  $75 ^-.12  =  625.     Ans.  $625. 

PROOF. — $625X-06=$37.50,  the  interest  for  1  year  at  the 
given  per  cent.,  and  $37.50X2=$75,  the  given  interest.  Hence, 

423*  To  find  the  principal,  when  the  interest,  rate  per  cent., 
and  time  are  given. 

.Divide  the  given  interest  by  the  interest  of  $1  for  the  given 
time  and  rate,  expressed  in  decimals  ;  and  the  quotient  will  be  the 
principal  required. 

Or,  make  the  interest  of  $1  for  the  given  time  and  rate,  the  numer- 
ator, and  100  the  denominator  of  a  common  fraction  ;  tlien  divide 
ttte  given  interest  by  this  fraction,  and  the  quotient  will  be  tlie  prin- 
cipal required.  (Art.  234.) 

13.  What  sum  must  be  put  at  7  per  cent,  interest,  to  gain  $63 
in  6  months? 

14.  What  sum  must  be  put  at  5  per  cent,  interest,  to  gain  $90 
in  4  months  ? 

15.  What   sum  must  be  invested  in  6  per  cent,  stock,  to  gain 
$300  in  6  months  ? 

QnsT. — 423.  When  the  interest,  rate  per  cent.,  and  time  arc  given,  how  is  the  jrinci 
pal  found  ? 

12* 


268  APPLICATIONS    OF  [SECT.  XII. 

16.  What  sum  must  be  invested  in  7  per  cent,  stock,  to  gain 
$560  in  one  year? 

17.  A  man  founded  a  professorship  with  a  salary  of  Si 000  a 
year :  what  sum  must  be  invested  at  7  per  cent,  to  produce  it  ?  • 

18.  What  sum  must  be  put  at  6  per  cent,  interest  to  pay  a 
salary  of  $1200  a  year  ? 

19.  What  sum  must  be  invested  in  5  per  cent,  stock  to  make  a 
simi-annual  dividend  of  $7 50  ? 

20.  A  man  bequeathed  his  wife  $1250  a  year:  what  sum  must 
be  invested  at  6  per  cent,  interest  to  pay  it  ? 

PROBLEM     IV. 

To  find  the  TIME,  tlie  principal,  the  interest,  and  the  rate  per 
cent,  being  given. 

21.  A  man  loaned  $200  at  6  per  cent.,  and  received  $42  inter- 
est :  how  long  was  it  loaned  ? 

Analysis. — The  interest  of  $200  at  6  per  cent,  for  1  year  is  $12. 
(Art.  404.)  Now,  since  $12  interest  requires  the  principal  1  year 
at  the  given  per  cent.,  $42  interest  will  require  the  same  princi^ 
pal  ^-f  of  1  year,  which  is  equal  to  3£  years.  (Art.  196.) 

Or,  we  may  reason  thus  :  If  $12  interest  requires  the  use  of  the 
given  principal  1  year,  $42  interest  will  require  the  same  prin- 
cipal as  many  years  as  $12  is  contained  times  in  $42,  And 
$42 -i- $12=3.5.  Ans.  3.5  years.  Hence,  , 

424.  To  find  the  time,  when  the  principal,  interest,  and  rate 
per  cent,  are  given. 

Divide  the  given  interest  by  the  interest  of  the  principal  at  the 
given  rate  for  1  year,  and  the  quotient  will  be  the  time  required. 

Or,  make  the  given  interest  the  numerator,  and  the  interest  of  the 
principal  for  1  year  at  tlie  given  rate  the  denominator  of  a  common 
fraction  ;  reduce  this  fraction  to  a  whole  or  mixed  number,  and  it 
vill  be  the  time  required. 

OBS.  If  the  quotient  contains  a  decimal  of  a  year,  it  should  be  reduced  to 
months  and  days.  (Art.  348.) 

QUKST.— 424.  When  the  principal,  interest,  and  rate  per  cent,  are  given,  how  is  the  time 
found  ?  Obs.  When  the  quotient  contains  a  decimal  of  a  year,  what  should  be  done  with  it  1 


ART.  424  ] 


INTEREST 


269 


22.  A  man  loaned  $765.50,  at  6  per  cent.,  and  received  $183.72 
interest :  how  long  was  it  loaned  ? 

23.  In  what  time  will  8850  gain  $29.75,  at  7  per  cent,  per 
•annum  ? 

24.  A   man  received   8136.75   for  the   use  of  $1820,  which 
was  C  per  cent,  interest  for  the  time :  what  was  the  time  ? 

25.  Tn  what  time  will  $6280  gain  $471,  at  5  per  cent,  interest? 
20.   li  >w  long  will  it  take  $100,  at  5  per  cent.,  to  gain  $100 

interest;  that  is,  to  double  itself? 

Operation.  The.  interest  of  $100  for  1  year,  at  5  per  cent., 

$5)^100  is  $5.  (Art.  404.) 

20  Ans.  20  years. 

PROOF. — $100  X  .05  X  20=$100,  the  given  principal.  (Art.  404.) 

TABLE, 

Showing  in  what  time  any  given  principal  will  double  itself  at  any  rate, 
from  1  to  20  per  cent.  Simple  Interest. 


Per  cent. 

Years. 

Per  cent. 

Years. 

Per  cent. 

Years. 

Per  cent.' 

Years. 

1 

100 

6 

101 

11 

9-rV 

16 

6| 

2 

50 

7 

14? 

12 

8* 

17 

Mf 

3 

33i 

8 

12* 

13 

VA 

18 

5f 

4 

25 

9 

iii 

14 

H 

19 

5A 

5 

20 

10 

10 

15 

ci 

20 

5 

27.  How  long  will  it  take  $365  to  double  itself,  at  6  per  cent.  ? 

28.  How  long  will  it  take  $1181  to  double  itself,  at  7  per  cent.  ? 

29.  In  what  time  will  $2365.24  double  itself  at  7  per  cent.? 

30.  In  what  time  will  $5640  double  itself,  at  10  per  cent.  ? 

31.  How  long  will  it  take  $10000  to  gain  $5000,  at  6  per  cent. 
Interest  ? 

32.  A  man  hired  $15000,  at  7  per  cent.,  and  retained  it  till  the 
principal  and  interest  amounted  to  $25000 :    how  long  did  he 
have  it  ? 

33.  A  man  loaned  his  clerk  $25000  to  go  into  business,  and 
agreed  to  let  him  have  it,  at  5  per  ct.,  till  it  amounted  tc  $60000 : 
how  long  did  he  have  it  ? 


270  COMPOUND  [SECT.  XIL 

COMPOUND   INTEREST. 

425*  Compound  Interest  is  the  interest  arising  not  only  from 
the  principal,  but  also  from  the  interest  itself,  after  it  becomes 
due. 

OBS.  Compound  Interest  is  often  called  interest  upon  interest.  When  inter- 
est is  paid  on  the  principal  only,  it  is  called  Simple  Interest. 

Ex.  1.  What  is  the  compound  interest  of  $842  for  4  years,  at 
6  per  cent.  ? 

Operation. 
$842.00  Principal. 

$842X.06  =      50.52   Int.  for  1st  year. 

~892.52  Amt.  for  1  year. 

$802.52 X. 00=     53.55  Int.  for  2d  year. 

946.07   Amt.  for  2  years. 

$946.07X.06  =     56.76  Int.  for  3d  year. 

1002.83  Amt.  for  3  years. 

$1002.83 X- 06=     00.17  Int.  for  4th  year. 

1063.00  Amt.  for  4  years. 

842.00  Prin.  deducted. 
Ans.  $221.00  Compound  int.  for  4  years. 

426.  Hence,  to  calculate  compound  interest. 

Cast  the  interest  on  the  given  principal  for  1  year,  or  the  specified 
time,  and  add  it  to  the  principal ;  then  cast  th#  interest  on  this 
amount  for  the  next  year,  or  specified  time,  and  add  it  to  the  prin- 
cipal as  before.  Proceed  in  this  manner  with  each  successive  year 
of  the  proposed  time.  Finally,  subtract  the  given  principal  from 
the  last  amount,  and  the  remainder  will  be  the  compound  interest. 

2.  What  is  the  compound  interest  of  $600  for  5  years,  at  7  per 
cent.?  Ans.  $241.53. 

3.  What  is  the  compound  int.  of  $1260  for  5  yrs.,  at  7per  cent.  ? 

4.  What  is  the  amount  of  $1535  for  6  yrs.,  at  6  percent,  com- 
pound interest  ? 

5.  What  is  the  amount  of  $4000  for  2  yrs.,  at  7  per  cent.,  paya« 
ble  semi-annually  ? 

QUEST.— 426.  How  is  compound  interest  calculated? 


ARTS.  425,426.] 


INTEREST . 


271 


TABLE, 

Sfwviins;  the  amount  of  $1,  or  £1,  at  3,  4,  5,  6,  and  7  per  cent,  compound 
interest,  for  any  number  of  years,  from  1  to  40. 


j  Yrs. 

M  prr  cent. 

4  per  cent.      1     5  per  cent. 

6  per  cent. 

7  per  cent. 

1. 

1.030,000 

1.040,000 

1.050,000 

1.060,000 

1.07,000 

2. 

1.060,900 

1.081,600 

1.102,500 

1.123,600 

1.14,490 

3. 

1.092,727 

1.124,864 

1.157,625 

1.191,016 

1.22,504 

4. 

1.125,509 

1.169,859 

1.215,506 

1.262,477 

1.31,079 

5. 

1.159,274 

1.216,653 

1.276,282 

1.338,226 

1.40,255 

6. 

1.194,052 

1.265,319 

1.340,096 

1.418,519 

1.50,073 

7. 

1.229,874 

1.315,932 

1.407,100 

1.503,630 

1.60,578 

8. 

1.266,770 

1.368,569  1   1.477,455 

1.593,848 

1.71,818 

9. 

1.304,773 

1.423,312 

1.551,328 

1.689,479 

1.83,845 

10. 

1.343,916 

1.480,244 

1.628,895 

1.790*848 

1.96,715 

11. 

1.384,234 

1.539,454 

1.710,339 

1.898,299 

2.10,485 

12. 

1.425,761 

1.601,032 

1.795,856 

2.012,196 

2.25,219 

13. 

1.468,534 

1.665,074 

1.885,649 

2.132,928, 

2.40,984 

14. 

1.512,590 

1.731,676 

1.979.932 

2.260,904 

2.57,853 

15. 

1.557,967 

1.800,944 

2.078,928 

2.396,558 

2.75,903 

16. 

1.604,706 

1.872,981 

2.182,875 

2.540,352 

2.95,216 

17. 

1.652,848 

1.947,900 

2.292,018 

2.692,773 

3.15,881 

18. 

1.702,433 

2.025,817 

2.406,619 

2.854,339 

3.37,293 

19. 

1.753,506 

2.106,849 

2.526,950 

3.025,600 

3.(>  1,652 

20. 

1.806,111 

2.191,123 

2.653,298 

3.207,135 

3.86,968 

21. 

1.860,295 

2.278,768 

2.785,963 

3.399,564 

4.14.056 

22. 

1.916,103 

2.3(.'9,919 

2.925,261 

3.603,537 

4.43,040 

23. 

1.973,587 

2.464,716 

3.071,524 

3.819,750 

4.74,052 

24. 

2.032,794 

2.563,304 

3.225,100 

4.048,935 

6.07,236 

25. 

2.093*,778 

2.665.836 

3.386,355 

4.291,871 

«.  42,743 

26. 

2.156,592 

2.772,470 

3.555,673 

4.549,383 

•S.80,735 

27. 

2.221,289 

2.883,369 

3.733,456 

4.822,346 

o.21,386 

28. 

2.287,928 

2.998,703  i  3.920,129 

5.111,687 

o.64,883 

29. 

2.356,566 

3.118,651   !  4.116,136 

5.418,388 

7.11,425 

30.  j    •  427,262 

3.243,398 

4.321.942 

5.743,491 

7.61,225 

31.  I  ^.500,080  '   3.373,133 

4.538,039 

6.088,101 

8.14,571 

32. 

2.575,083      3.508,059      4.764,941 

6.453,386 

8.71.527 

33. 

2.652,335      3.648,381      5.003,189      6.840,590 

9.32.533 

34. 

2.731,905  j  3.794,316      5.253,348      7.251,025 

9.97,811 

35. 

2.813,862      3.946,089      5.516,015      7.686.087 

10.6,765 

36. 

2.898,278      4.103,933'     5.791,816      8.147,252 

11.4,239 

37. 

2.985,227 

4.268,090      6.081,407      8.636,087 

12.2,236 

38. 

3.074,783 

4.438,813      6.385,477      9.154,252 

13  0,792 

39. 

3.167,027  I  4.616,366      6.704,751   !   9.703,507       I'S  9,948 

40. 

3.262,038      4.801,021  j  7.039,989  ;   10.285,72       ^.9*4 

DISCOUNT  [SECT.  XIL 

427.  To  calculate  compound  interest  by  the  preceding  table. 

Find  the  amount  of  $1  or  £l  for  the  given  number  of  years  by 
the  table,  multiply  it  by  the  given  principal,  and  the  product  will 
be  the  amount  required.  Subtract  the  principal  from  the  amount 
thus  found,  and  the  remainder  will  be  the  compound  interest. 

6.  What  is  the  compound  interest  of  $500  for  15  years,  at  6 
per  cent.  ?  What  is  the  amount  ? 

Operation. 
$2.396558  Amt.  of  $1  for  15  yrs.  by  Table. 

500  The  given  principal. 
$1198.279000  Amt.  required. 
$500  Principal  to  be  subtracted. 

$698.279  Interest  required. 

7.  What  is  the  amount  of  $960  for  10  yrs.,  at  7  per  ct.  ? 

8.  What  is  the  amount  of  $1000  for  9  yrs.,  at  5  per  ct.  ? 

9.  What  is  the  compound  int.  of  $1460  for  12  yrs.,  at  4  per  ct.  ? 

10.  What  is  the  compound  int.  of  $2500  for  15  yrs.,  at  6  per  ct.  ? 

11.  What  is  the  amount  of  $5000  for  20  yrs.,  at  6  per  ct.  ? 

12.  What  is  the  amount  of  $10000  for  40  yrs.,  at  7  per  ct.  ? 

DISCOUNT. 

428*  DISCOUNT  is  the  abatement  or  deduction  made  for  the 
payment  of  money  before  it  is  due.  For  example,  if  I  owe  a  man 
$100,  payable  in  one  year  without  interest,  the  present  worth  of 
the  note  is  less  than  $100 ;  for,  if  $100  were  put  at  interest  for 
1  year,  at  6  per  cent.,  it  would  amount  to  $106  ;  at  7  per  cent., 
to  $107,  &c.  In  consideration,  therefore,  of  the  preseni  payment 
of  the  note,  justice  requires  that  he  should  make  some  ciiatement 
from  it.  This  abatement  is  called  Discount. 

429*  The  present  worth  of  a  debt  payable  at  some  future  time 
without  interest,  is  that  sum  which,  being  put  at  legal  interest, 
will  amount  to  the  debt,  at  the  time  it  becomes  due. 


QUEST. — 428.  What  is  discount  1    429   What  '•*  the  present  worth  of  a  debt,  payable  at 
some  future  time,  without  interest  7 


ARTS.  427-430.]  DISCOUNT.  273 

Ex.  1.  What  is  the  present  worth  of  $756,  payable  in  1  year 
and  4  months,  without  interest,  when  money  is  worth  6  per  cent, 
per  annum  ? 

Analysis. — The  amount,  we  have  seen,  is  the  sum  of  the  prin- 
cipal and  interest.  (Art.  399.)  Now  the  amount  of  $1  for  1  year 
and  4  months,  at  6  per  cent.,  is  81.08;  (Art.  404;)  that  is,  the 
amount  is  +$%  of  the  principal  $1.  The  question  then  resolves 
itself  into  this  :  $756  is  Ig-g-  of  what  principal  ?  If  $756  is  -f  ft-g 
of  a  certain  sum,  T-J-g-  is  TUT  of  $756  ;  now  $756-Hl08=$7,  and 
-^=$7X100,  which  is  $700. 

Or,  we  may  reason  thus :  Since  $1.08  (amount)  requires  $1 
principal  for  the  given  time,  $756  (amount)  will  require  as  many 
dollars  as  81.08  is  contained  times  in  $756 ;  and  $756-i-$1.08  = 
$700,  the  same  as  before. 

PROOF. — $700  X  .08=$56,  interest  for  1  year  and  4  months  ;  and 
$7004-56— $756,  the  sum  whose  present  worth  is  required.  Hence, 

43 O.  To  find  the  present  worth,  of  any  sum,  payable  at  a  future 
time  without  interest. 

first  find  the  amount  of  $1  for  the  time,  at  the  given  rate,  as 
in  simple  interest  j  then  divide  the  given  sum  by  this  amount,  and 
the  quotient  will  be  the  present  worth.  (Art.  404.) 

The  present  worth  subtracted  from  the  debt,  will  give  the  true 
discount. 

OBS.  This  process  is  often  classed  among  the  Problems  of  Interest,  in  which 
the  amount,  (which  answers  to  the  given  sum  or  debt,)  the  rate  per  cent.,  and 
the  time  are  given,  to  find  the  principal,  which  answers  to  the  present  worth. 

2.  What  is  the  present  worth  of  $424.83,  payable  in  4  months, 
when  money  is  worth  6  per  cent.  ?     What  is  the  discount  ? 
Solution. — $424.83-^$1.02=$416.50,  Present  worth. 
And  $424.83 — $416.50=$8.33,     Discount. 

3  What  is  the  present  worth  of  $1000,  payable  in  1  year, 
•when  the  rate  of  interest  is  7  per  cent.  ? 

4.  What  is  the  present  worth  of  $1645,  payable  in  1  year  and 
6  months,  when  the  rate  of  interest  is  7  per  cent.  ? 


UUKST. — 430.  How  do  you  find  the  present  worth  of  a  det  1    HoW  find  the  discount? 


274  BANK    DISCOUNT.  [SECT.  All. 

5.  What  is  the  discount  on  a  note  for  82300,  payable  in  6 
months,  when  the  rate  of  interest  is  8  per  cent.  ? 

6.  What  is  the  discount,  at  6  per  cent.,  on  $4260,  payable  in 
4  months  ? 

7.  What  is  the  present  worth  of  a  note  for  $4800,  due  in  3 
months,  when  the  rate  of  interest  is  0  per  cent.  ?    . 

8.  What  is  the  present  worth  of  a  draft  for  $6240,  payable  ia 
1  month,  when  the  rate  of  interest  is  6  per  cent.  ? 

9.  A  man  sold  his  farm  for  $3915,  payable  in  2-£  years :  what 
is  the  present  worth  of  the  debt,  at  6  per  cent,  discount  ? 

10.  What  is  the  present  worth  of  a  draft  of  $10000,  payable  at 
30  days  sight,  when  interest  is  6  per  cent,  per  annum  ? 

11.  What  is  the  difference  between  the  discount  of  $8000  for 
1  year,  and  the  interest  of  $8000  for  1  year,  at  7  per  cent.  ? 

BANK  DISCOUNT. 

431*  A  Bank,  in  commerce,  is  an  institution  established  for 
the  safe  keeping  and  issue  of  money,  for  discounting  notes,  deal- 
ing in  exchange,  &c. 

OBS.  1.  There  are  three  kinds  of  banks,  viz:  banks  of  deposit,  discount,  and 
circulation. 

A  bank  of  deposit  receives  money  to  keep,  subject  to  the  order  of  the  de- 
positor. This  was  the  primary  object  of  these  institutions. 

A  bank  of  discmmt  is  one  which  loans  money,  or  discounts  notes,  drafts, 
and  bills  of  exchange. 

A  bank  of  circulation  issues  bills,  or  notes  of  its  own,  which  are  redeem- 
able in  specie,  at  its  place  of  business,  and  thus  become  a  circulating  medium 
of  exchange.  Banks  of  this  country  generally  perform  the  three-fold  office  of 
deposit,  discount,  and  circulation. 

2.  The  affairs  of  a  bank  are  managed  by  a  board  of  directors,  chosen  annu- 
ally by  the  stockholders.  (Art.  392.  Obs.)    The  directors  appoint  a  president  and 
cashier,  who  sign  the  bills,  and  transact  the  ordinary  business  of  the  bank. 

A  teller  is  a  clerk  in  a  bank,  who  receives  and  pays  the  money  on  checks. 
A  check  is  an  order  for  money,  drawn  on  a  banker,  or  t  le  casl  icr,  by  a  de- 
positor, payable  to  the  bearer. 

3.  Banks  originated  in  Italy.     The  first  one  was  established  m  Venice,  m 
1171,  called  the  Bank  of  Venice. 

QUEST. — 431.  What  is  a  bank  ?     Obs.  Of  how  many  kinds  are  banks  ? 


ARTS.  431-433.]          BANK  DISCOUNT.  275 

432.  It  is  customary  for  Banks,  in  discounting  a  note  or 
draft,  to  deduct  in  advance  the  legal  interest  on  the  given  sum 
from  the  time"  it  is  discounted  to  the  time  it  becomes  due.  Hence, 

Bank  discount  is  the  same  as  simple  interest  paid  in  advance. 
Thus,  the  lank  discount  on  a  note  of  $106,  payable  in  1  year,  at  .8 
per  cent.,  is  $6.36,  while  the  true  discount  is  but  $6.  (Art.  430.) 

OBS.  1  The  difference  between  bank  discount  and  true  discount,  is  the  inter- 
est of  tht  true  discount  for  the  given  time.  On  small  sums  for  a  short  period 
his  difference  is  trifling,  but  when  the  sum  is  large,  and  the  time  for  vhich  it 
is  discounted  is  long,  the  difference  is  worthy  of  notice. 

2.  Taking  legal  interest  in  advance,  according  to  the  general  rule  of  law,  a 
'usury.  An  exception  is  generally  allowed,  however,  in  favor  of  notes,  drafts, 
&c.,  which  are  payable  in  less  than  a  year. 

The  Safety  Fund  Banks  of  the  State  of  New  York,  though  the  legal  rate 
of  interest  is  7  per  cent.,  are  not  allowed  by  their  charters  to  take  over  6  per 
cent,  discount  in  advance  on  notes  and  drafts  which  mature  within  63  days 
from  the  time  they  are  discounted.* 

Banks  charge  interest  for  the  three  days  grace. 

CASE    I. 

12.  What  is  the  bank  discount  on  a  note  for  $850.20  for  6 
months,  at  6  per  cent.  ?  What  is  the  present  worth  of  the  note  ? 

Operation. 
$850.20  Principal. 

.03  05  Int.  $1  for  6  mo.  3  ds.  grace. 
425100 
25  5060 

$25.9311  00  Bank  discount. 
And  $850.20— $25.93 =$824.27,  Present  worth.     Hence, 

433*  To  find  the  bank  discount  on  a  note  or  draft. 

Cast  tlie  interest  on  the  face  of  tlie  note  r  draft  for  three  days 
more  than  the  specified  time,  and  the  result  will  le  tlie  discount. 

The  discount  subtracted  from  the  face  of  the  note,  ivill  give  th» 
present  worth  o*  proceeds  of  a  note  discounted  at  a  bank. 

QUEST. — 432.  How  do  banks  usually  reckon  discount?  What  then  is  bank  discount! 
Oft*.  What  is  the  difference  between  bank  discount  and  true  discount?  Is  this  difference 
worth  noticing  ?  How  is  taking  interest  in  advance  generally  regarded  in  law  1  What 
exception  to  this  rule  is  allowed  ? 

•  Revised  Statutes  of  New  York.  (3d  edition,)  Vol.  I.  p.  741. 


276  BANK    DISCOUNT.  [SECT.  XII. 

Note.— Intel  est  should  be  computed  for  the  three  days  grace  in  each  of  the 
following  examples. 

1  ±.  What  is  the  bank  discount  on  a  note  for  $^65,  payable  in 
6  months,  at  6  per  cent.  ? 

15.  What  is  the  bank  discount  on  a  note  for  $972,  payable  in 
4  months,  at  5  per  cent.  ? 

16.  What  is  the  bank  discount  on  a  note  for  $1492,  payable  in 
3  months,  at  7  per  cent.  ? 

17.  What  is  the  bank  discount  on  a  draft  of  $028,  payable  at 
60  days  sight,  at  5  per  cent.  ? 

18.  What  is  the  present  worth  of  $2135,  payable  in  8  months, 
at  7  per  cent.  ? 

19.  What  is  the  present  worth  of  a  note  for  $2790,  payable  in 
1  month,  discounted  at  C  per  cent,  at  a  bank  ? 

20.  What  is  the  bank  discount,  at  5i  per  cent.,  on  a  draft  of 
$1747,  payable  at  90  days  sight? 

21.  What  is  the  bank  discount,  at  4£  per  cent.,  on  a  draft  of 
$3143,  payable  in  4  months? 

22.  What  is  the  bank  discount  on  $5126.63,  payable  in  30  days, 
at  8  per  cent.  ? 

23.  What  is  the  bank  discount  on  $3841.27,  payable  hi  60  days, 
at  6£  per  cent.  ? 

24.  What  is  the  present  worth  of  a  note  for  $6721,  payable  in 
10  months,  discounted  at  6  per  cent,  at  a  bank? 

25.  What  is  the  present  worth  of  a  note  for  $1500,  payable  in 
1 2  days,  at  7  per  cent,  discount  ? 

26.  What  is  the  bank  discount  on  $10000,  payable  in  45  days, 
at  6  per  cent.  ? 

27.  What  is  the  bank  discount  on  $25260,  payable  in  90  days, 
at  7  per  cent.  ? 

28.  What  is  the  difference  between  the  true  discount  and  bank 
discount  on  $5000  for  10  years,  at  6  per  cent.  ? 

CASE    II 

29.  A  man  wishes  to  make  a  note  payable  in   1  year,  at  6 
per  cent.,  the  present  worth  of  which,  if  discounted  at  a  bank, 
shall  be  just  $200  :  for  what  sum  must  the  note  be  made  ? 


A.RT.  434.]  BANK    DISCOUNT.  277 

Analysis. — The  present  wortli  of  8l,  payable  in  1  year,  at  6 
per  cent,  discount,  is  100  cts. — G  cts. =94  cts. ;  that  is,  the  present 
worth  is  -^jfV  of  the  principal  or  sum  discounted.  The  question 
then  resolves  itself  into  this  :  $5200  (present  worth)  is  -f^f.  of 
what  sum?  Now,  if  $200  is  -ftfa  of  a  certain  sum,  -rJnr  is  -fa 
of  1200;  and  8200-^94=82.12766,  and  iH— 82.12766X  100, 
which  is  8212.766.  Ans. 

Or,  we  may  reason  thus:  Since  94  cents  present  worth  requires 
$1,  (100  cents)  principal,  or  sum  to  be  discounted  for  the  ghen 
time,  8200  present  worth  will  require  as  many  dollars,  as  94  cents 
is  contained  times  in  $200 ;  and  8200^-8-94=8212.766, 

PROOF. — $212.766X-06=8l2.7659,  the  bank  discount  for  1 
year;  and  8212.766—812.7659=8200,  the  given  sum.  Hence, 

434.  To  find  what  sum,  payable  in  a  specified  time,  will 
produce  a  given  amount,  when  discounted  at  a  bank,  at  a  giveA 
per  cent. 

Divide  the  given  amount  to  be  raised  by  the  present  worth  of  8l, 
for  the  time,  at  the  given  rate  of  bank  discount,  and  the  quotient 
will  be  the  sum  required  to  be  discounted. 

30.  How  large  must  I  make  a  note  payable  in  6  months,  to  raise 
$400,  when  discounted  at  7  per  cent,  bank  discount? 

31.  What  sum  payable  in  4  months  must  be  discounted  at  a 
bank,  at  5  per  cent.,  to  produce  8950  ? 

32.  What  sum  payable  in  60  days,  will  produce  $1236,  if  dis- 
counted at  a  bank,  at  8  per  cent.  ? 

33.  For  what  sum  must  a  note  be  drawn,  payable  in  34  days, 
the  avails  of  which,  at  6  per  cent.,  bank  discount,  will  be  82500  ? 

34.  For  what  sum  must  a  note  be  drawn,  payable  in  90  days, 
so  that  the  avails,  at  7  per  cent,  bank  cfiscount,  shall  be  83755  ? 

35.  A  man  bought  a  farm  for  $4208  cash :  how  large  a  note 
payable  in  4  months,  must  he  take  to  bank  to  raise  the  money  at 
6  per  cent,  discount  ? 

QrusT. — 134.  How  find  what  sum,  payable  in  a  given  time,  will  »roduce  .'i  given  amount* 
at  a  given  per  cent.,  bank  discount  ? 


278  INSURANCE.  [SECT.  XII. 

36  A  man  wishes  to  obtain  $63240  from  a  bank  at  6  per  cent, 
discount :  how  large  must  he  make  his  note,  payable  in  1  month 
and  15  days? 

37.  What  sum  payable  in  8  months,  if  discounted  at  a  bank,  at 
6  per  sent.,  will  produce  $10000? 

38.  What  sum  payable  in  4  months,  will  produce  $50000,  if 
discounted  at  7  per  cent,  at  a  bank  ? 

39.  A  man  received  $46250  as  the  avails  of  a  note,  payable  in 
60  days,  discounted  at  a  bank  at  5  per  cent. :  what  was  the  face 
of  the  note  ? 

40.  A  merchant  wished  to  pay  a  debt  of  $8246  at  a  bank,  by 
getting  a  note  payable  in  30  days  discounted,  at  8  per  cent. :  how 
large  must  he  make  the  note  ? 

INSURANCE. 

435.  INSURANCE  is  security  against  loss  or  damage  of  prop- 
erty by  fire,  storms  at  sea,  and  other  casualties.  This  security 
is  usually  effected  by  contract  with  Insurance  Companies,  who, 
for  a  stipulated  sum,  agree  to  restore  to  the  owners  the  amount 
insured  on  their  houses,  ships,  and  other  property,  if  destroyed 
or  injured  during  the  specified  time  of  insurance. 

OBS.  1.  Insurance  on  ships  and  other  property  at  sea  is  sometimes  effected 
by  contract  with  individuals.  It  is  then  called  out-door  insurance. 

2.  The  insurers,  whether  an  incorporated  company  or  individuals,  are  often 
termed  Underwriters. 

436*  The  written  instrument  or  contract  is  called  the  Policy. 

The^s-wm  paid  for  insurance  is  called  the  Premium. 

The  premium  paid  is  a  certain  per  cent,  on  the  amount  of  prop- 
erty insured  for  1  year,  or  during  a  voyage  at  sea,  or  other  spe- 
cified time  of  risk. 

OBS.  1.  Rates  of  insurance  on  dwelling-house?  and  furniture,  stores  and 
goods,  shops,  manufactories,  &c.,  vary  from  -f  to  2  per  cent,  per  annum  on 
the  sum  insured,  according  to  the  exposure  of  the  property  and  the  difficulty 
of  moving  the  goods  in  case  of  casualty.  It  is  a  rule  with  most  Insurance 

QUEST. — 435.  What  is  Insurance?  Obs.  When  insurance  is  effected  with  individuals, 
what  is  it  called  ?  What  are  the  insurers  sometimes  called  ?  436.  What  is  meant  by  the 
policy?  The  premium? 


ARTS.  435-437.]  INSURANCE.  279 

Companies  not  to  insure  more  than  two  thirds  of  the  valuo  of  a  building,  or 
goods  on  land. 

2.  Coasting  vessels  are  commonly  insured  by  the  season  or  year.     In  time 
of  peace,  the  rate  varies  from  4  to  7£  per  cent,  per  annum;  it.  time  of  war  it  is 
much  higher.     Whale  ships  are  generally  insured  for  the  voyage,  ttt  a  rate 
varying  from  5  to  8  per  cent,  on  the  sum  insured. 

3.  When  the  general  average  of  loss  is  less  than  5  per  cent.,  the  underwriter* 
are  not  liable  for  its  payment. 

CASE     I. 

437.  To  compute  Insurance  for  1  year,  or  a  specified  time. 
Multiply  the  sum  insured  by  the  given  rate  per  cent.,  as  in  inter- 
est. (Art.  404.) 

Ex.  1.  A  man  effected  an  insurance  on  his  house  for  $500,  at 
\\  per  cent,  per  annum :  how  much  premium  did  he  pay  ? 

Solution. — $1500X.0125  (the  rate)=$l 8.75.  Ans. 

2.  What  is  the  premium  for  insuring  a  store  to  the  amount  of 
$2760,  at  |  per  cent.  ? 

3.  What  premium  must  I  pay  for  insuring  a  quantity  of  goods, 
worth  '156280,  from  New  York  to  Liverpool,  at  1£  per  cent.  ? 

4.  What  is  the  annual  premium  for  insuring  a  stock  of  goods, 
worth  $10200,  at  f  per  cent.  ? 

5.  What  is  the  annual  premium  for  insuring  a  coasting  vessel, 
worth  $1600,  at  6|  per  cent.? 

6.  A  bookseller  shipped  a  quantity  of  books,  valued  at  $4700, 
/rom  Boston  to  New  Orleans,  at  1^  per  cent,  insurance :    what 
amount  of  premium  did  he  pay  ?  • 

7.  A  merchant  shipped  a  cargo  of  flour,  worth  $45000,  from 
New  York  to  Liverpool,  at  2  per  cent. :  how  much  premium  did 
he  pay  ? 

8.  What  is  the  insurance  on  a  cargo  of  teas,  worth  $75000, 
from  Canton  to  Philadelphia,  at  2£  per  cent.  ? 

9.  What  is  the  annual  insurance  on  a  factory,  worth  $65000, 
at  -f  per  cent.  ? 

10.  A   powder  mill  was  insured  for  $1945,  at  12^  per  cent.: 
what  was  the  annual  premium  ? 


QXJKST.— 437.  How  is  insurance  computed  for  1  year  or  a  specified  time  ? 


280  INSURANCE.  [SECT.  XII 

11.  A  ship  embarking  on  an  exploring  expedition,  was  insured 
for  $45360,  at  8£  per  cent,  per  annum :  what  did  the  insurance 
amount  to  in  5  years  ? 

12.  A  policy  of  insuiance  for  $45000  was  obtained  on  a.  whale 
ship,  at  7£  per  cent,  for  the  voyage :  what  was  the  amount  paid 
for  insurance  ? 

CASE    II. 

1 "  If  a  man  pays  $16  annually  for  insuring  $800  on  his  shop, 
What  pei  cent,  does  he  pay  ? 

Analysis. — If  $800,  the  amount  insured,  costs  $16  premium, 
#1  will  cost  rJ-o  of  $16  ;  and  $16-^800=.02  ;  which  is  2  per  cent. 
PROOF. — $800 X- 02 =$16,  the  premium  paid.     Hence, 

43  8,  To  find  the  rate  per  cent,  when  the  sum  insured  and  the 
annual  premium  are  given. 

Divide  the  given  premium  by  the  sum  insured^  and  the  quotient 
will  be  the  rate  per  cent,  required. 

Note. — This  case  is  similar  in  principle  to  Problem  II.  in  Interest. 

14.  If  a  man  pays  $60  annually  for  insuring  $2400  on  his 
house,  what  per  cent,  does  it  cost  him  ? 

15.  A  merchant  pays  $200  per  annum  for  insuring  $8000  on  his 
goods  :  what  per  cent,  does  he  pay  ? 

16.  A  grocer  paid  $122.50    premium  on   a    cargo    of   flour, 
worth  $12250,  from  Charleston  to  Portland:  what  per  cent,  did 
he  pay  ? 

17.  An  importer  paid  $350  insurance  on  a  quantity  of  cloths, 
worth  $28000,  from  Havre  to  New  York :  what  per  cent,  did  he 
pay? 

CAS  E    III. 

18.  A  man  pays  $45  annually  for  insuring  his  library,  which  is 
8  per  cent,  on  the  amount  of  his  policy  :  what  is  the  sum  insured  ? 

Analysis. — Since  3  cents  will  insure  $1  at  the  given  rate,  for  a 
year,  $45  will  insure  as  many  dollars  as  3  cents  are  contained 
times  in  $45  ;  and  $45 -f-. 03 =$1500.  Ans. 

PROOF. — $1500X.03=$45,  the  given  premium.     Hence, 


ARTS.  438,  439.]  INSURANCE.  281 

439*.  To  find  the  sum  insured  when  the  premium  and  thfc 
rute  per  cent,  are  given. 

Divide  the  given  premium  by  the  rate  per  cent.,  expressed  in  deci- 
mals, and  the  quotient  will  be  the  stim  insured. 

Note. — This  case  is  similar  in  principle  to  Problem  III.  in  Interest. 

19.  An  inr  porter  paid  $650  prenmim  on  goods  from  Hamburgh 
to  New  York,  which  was  1-J-  per  cent,  on  the  amount  insured : 
li  >w  much  did  he  insure  ? 

20.  A  merchant  paid  $1640  premium  on  goods  from  Philadel- 
phia to  Constantinople,  which  was  2£  per  cent,  on  the  worth  of 
the  goods  insured  :  how  much  did  he  insure  ? 

21.  A  premium  of  $48*7.50  was  paid  on  a  cargo  of  cotton  from 
New  Orleans  to  Liverpool,  which  was  •$•  per  ce»t.  on  its  value : 
what  amount  was  insured  on  the  cargo  ? 

22.  When  the  rate  of  insurance  if  !-£  per  cent.,  what  sum  can 
you  get  insured  for  $860  premium  ? 

23.  At  •§•  per  cent,  per  annum,  what  amount  can  a  man.  get  in- 
sured on  his  house  and  furniture  for  $20.50  per  annum? 

CASE    IV. 

To  find  what  sum  must  be  insured  on  any  given  property,  so 
that,  if  destroyed,  its  value  and  the  premium  may  both  be  recov- 
ered. 

24.  If  a  man  owns  a  vessel  worth  $1920,  what  sum  must  he  get 
insured  on  it,  at  4  per  cent.,  so  that  if  wrecked,  he  may  recover 
both  the  value  of  the  vessel  and  the  premium  ? 

Analysis. — It  is  plain,  when  the  rate  of  insurance  is  4  per  cent, 
on  a  policy  of  $1,  or  100  cents,  the  owner  would  receive  but  96 
cents  towards  his  loss ;  for,  he  has  paid  4  cents  for  insurance. 
Since  therefore  the  recovery  of  96  cents  requires  $1  to  be  insured, 
the  recovery  of  $1920  will  require  as  many  dollars  to  be  insured 
as  96  cents  is  contained  times  in  $1920;  and  $1920~.96  = 
$2000.  Ans. 

PROOF. — $2000X-04=$80,  the  premium  paid,  and  $2000 — 
$80 =$19 20,  the  value  of  the  vessel. 


. 


282  LIFE     INSURANCE.  [SuCT.  XII 

44O.  Hence,  to  find  what  sum  must  be  insured  on  a  given 
amount  of  property,  so  that  if  destroyed,  both  the  value  of  the 
property  and  the  premium  may  be  recovered. 

Subtract  the  rate  per  cent,  from  $1,  then  divide  the  value  of  the 
property  insured  by  the  remainder,  and  the  quotient  will  be  the  sum 
to  be  insured. 

25.  What  sum  must  be  uisured  on  property  worth  &8240,  at 
1J  per  cent.,  so  that  the  owner  may  suffer  no  loss  if  the  property 
ft)  destroyed  ? 

26.  What  sum  must  be  insured  on  $13460,  at  3  per  cent.,  in 
order  to  cover  both  the  premium  and  property  insured  ? 

27.  If  I  send  an  adventure  to  the  Sandwich  Islands  worth 
$25000,  what  sum  must  I  get  insured,  at  7£  per  cent.,  that  I  may 
sustain  no  loss  in  case  of  a  total  wreck  ? 

LIFE   INSURANCE. 

44 1»  A  LIFE  INSURANCE  is  a  contract  for  the  payment  of  a 
certain  sum  of  money  on  the  death  of  an  individual,  in  considera- 
tion of  a  stipulated  sum  paid  down,  or,  more  commonly,  of  an 
annual  premium,  to  be  continued  during  the  life  of  the  assured. 

The  average  duration  of  human  life  is  often  called  the  Expecta- 
tion of  Life.  This  is  different  in  different  countries,  but  it  may  be 
determined  with  great  accuracy  in  any  given  country,  by  calcula- 
tions founded  on  the  register  of  births  and  deaths  in  that  country. 

OBS.  At  birth,  the  expectation  of  life,  according  to  the  Carlisle  Table,  is 
38.72  y. ;  at  5,  it  is  51.25  y. ;  at  10,  it  is  48.82  y. ;  at  15,  it  is  45  y. ;  at  20,  it 
is  41.46  y. ;  at  25,  it  is  37.86  y. ;  at  30,  it  is  34.34  y. ;  at  35,  it  is  31  y. ;  at  40, 
it  is  27.61  y. ;  at  45,  it  is  24.46  y. ;  at  50,  it  is  21.11  y. ;  at  55,  it  is  17.58  y. ; 
at  60.  it  is  14.34  y. ;  at  65,  it  is  11.79  y. ;  at  70,  it  is  9.19  y. ;  at  75,  it  is  7.01  y. ; 
at  80^  it  is  5.51  y. ;  at  85,  it  is  4.12  y. ;  at  90,  it  is  3.28  y. ;  at  100,  it  is  2.28  y. 

442*  The  premium  paid  for  life  insurance,  like  that  for  other 
insurance,  is  calculated  at  a  certain  per  cent,  on  the  amount  in- 
sured.    The  per  cent,  varies  according  to  the  age  and  employmen 
of  the  assured,  and  the  time  embraced  in  the  policy. 

UCBST.— 441.  What  is  Life  Insurance?     What  is  meant  by  the  expectation  cf  lift  t 
442.  How  is  Life  insurance  calculated  1 

*  See  Registers  of  London,  Breslau,  Northampton,  &c. 


ARTS.  440-443.]         PROFIT  AND  LOSS.  283 

OBS.  1.  At  the  age  of  21  years,  the  per  cent,  on  a  policy  for  life  is  from  1-^ 
to  S2-jp  per  cent,  per  annum  on  the  sum  insured ;  tor  7  years,  it  is  from  -£-£  to 
!-£  per  cent,  per  annum ;  for  1  year,  from  -fc  to  1-f  per  cent. 

At  30,  on  a  policy  for  life,  it  is  from  2if^-  to  2-,^  per  cent,  per  annum ;  for 
7  years,  from  \.-^~  to  l-^  per  cent. ;  for  1  year,  from  l-^  to  1-^  per  cent. 

At  40,  on  a  policy  for  life,  it  is  from  3fV  to  3^  per  cent.;  for  7  years,  from 
1  fc  to  2-j20-  per  cent. ;  for  1  year,  from  1-f  to  2^-  per  cent. 

At  50,  on  a  policy  for  life,  it  is  from  4-^  to  4-,^  per  cent. ;  for  7  years,  from 
J^  to  3-,^  per  cent. ;  for  1  year,  from  1-^  to  2-^  per  cent. 

At  60,  on  a  policy  for  life,  it  is  from  6-fc  to  7  per  cent. ;  for  7  years,  from 
IY^  to  5  per  cent. ;  for  I  year,  from  3  ft -  to  4-fo  per  cent. 

28.  A  -young  man,  at  the  age  of  21  years,  effected  an  insurance 
for  $1500  for  life,  at  2-fV  per  cent. :  what  was  the  annual  premium  ? 

Am.  $31.50. 

29.  A  man,  at  the  age  of  30,  effected  a  life  insurance  for  $2700, 
for  7  years,  at  1W  per  cent. :  what  was  the  premium  ? 

30.  At  60  years  of  age,  a  man  effected  a  life  insurance  for  1  yea/ 
for  $5750,  at  6£  per  cent. :  how  much  premium  did  he  pay  ? 

31.  At  40  years  of  age,  a  man  effected  an  insurance  for  $10000 
for  life,  at  3-£  per  cent,  per  annum  ;  he  lived  till  he  was  75  years 
old  :  which  was  the  larger,  the  sum  paid  for  insurance,  or  the  sum 
insured  ? 

PROFIT   AND    LOSS. 

44*1.  PROFIT  and  Loss  in  commerce,  signify  the  sum  gained 
or  lost  in  ordinary  business  transactions.  They  are  reckoned  at 
a  certain  per  cent,  on  the  purchase  price,  or  sum  paid  for  the  arti- 
cles under  consideration. 

CASE    I . 

To  find  the  AMOUNT  of  profit  or  loss,  the  purchase  price  and 
rate  per  cent,  being  given. 

Ex.  1.  A  grocer  bought  a  lot  of  flour  for  $84,  and  sold  it  for 
7  per  cent,  profit :  how  much  did  he  make  by  his  bargain  ? 

QUEST. — 443.  What  is  meant  by  profit  and  loss  ?    How  are  they  reckoned  ?    444.  Ham 
h  the  amount  of  profit  or  loss  found,  when  the  cost  and  rate  per  cent,  are  given  ? 
T.H. 


284  PROFIT    AND    LOSS.  [SECT.   XII. 

Analysis. — Since  lie  gained  7  per  cent,  on  the  cost  of  the  flour, 
he  must  have  gained  -j-fo-  of  $84.  Now  -rh>  of  $S4  is  -£fa,  and 
•j-for  is  7  times  as  much,  which  is  .-f-^f =$5.88.  Ans. 

Or  thus  :  If  $1  (100  cents)  gain  7  cents,  $84  will  gain  84  times 
as  much  ;  and  $84x.07=$5.88,  the  same  as  before.  Hence, 

444*  To  find  the  amount  of  profit  or  loss,  when  the  purchase 
price  and  rate  per  cent,  are  given. 

Multiply  the  purchase  price  by  the  given  per  cent,  as  in  percent- 
aye  j  and  the  product  will  be  the  amount  gained  or  lost  by  the  trans- 
action. (Art.  388.) 

OES.  In  order  to  obtain  the  exact  profit  and  loss  in  mercantile  ope/ations,  it 
is  manifest  that  the  interest  on  the  cost  or  purchase  price  of  the  goods,  during 
the  time  they  have  been  on  ha-nd,  also  for  the  time  before  payment  is  received 
should  be  taken  into  consideration. 

2.  If  I  buy  a  piece  of  broadcloth  for  $120,  and  after  keeping  it 
6  months,  sell  it  at  8  per  cent,  advance  on  6  months  credit,  how 
much  shall  I  gain  if  I  pay  7  per  cent,  for  the  money  invested  ? 

Ans.  $1.20. 

3.  If  I  buy  a  farm  for  $1740,  and  sell  it  8  per  cent,  less  than 
cost,  how  much  do  I  lose?  Ans.  $139.20. 

4.  If  you  buy  a  house  for  $2180,  and  sell  it  at  10  per  cent, 
advance,  how  much  will  you  gain  by  your  bargain  ? 

5.  A  merchant  bought  goods  amounting  to  $3400,  and  retailed 
them  at  20  per  cent,  profit :  how  much  did  he  make  ? 

6.  A  grocer  bought  a  lot  of  flour  for  $6235,  and  sold  it  15  per 
cent,  less  than  cost:  what  was  his  loss? 

7.  A  speculator  bought  a  quantity  of  cotton  for  $24850,  and 
sold  it  at  5-£  per  cent,  advance :  how  much  did  he  make  by  the 
operation  ? 

8.  A  man  bought  a  block  of  stores  for  $58246,  and  sold  them 
at  118  percent,  advance  :  how  much  did  he  gain  ? 

9.  A  man  bought  wild  land  amounting  to  $125000,  and  after 
keeping  it  10  years,  sold  it  at  50  per  cent,  advance:  allowing 
money  to  be  worth  6  per  cent.,  did  he  make  rr  lose  by  the  oper 
ation ;  and  how  much  ? 


ARIS.  444,  i45.]         PROFIT  AND  LOSS.  285 

CASE    II. 

To  find  how  an  article  must  be  sold  to  gain  or  lose  a  specified 
per  cent.,  the  cost  being  given. 

10.  A  man  bought  a  building  lot  for  $625,  and  afterwards  sold 
it  so  as  to  gain  10  per  cent. :  how  much  did  he  sell  it  for  ? 

-  Operation.  Since  he  gained  10  per  cent.,  it  ia 

$625  purchase  price.       obvious  he  sold  it  for  the  purchase 

.10  percent,  profit.       price  together  with  10  per  cent,  of 

$62.50  profit.  that    price.     We  therefore  find   10 

$687.50  selling  price.  per  cent,  on  the  cost,  and  add  it  to 

itself.  (Art.  388.) 

11.  A  man  bought  a  small  house  for  $840,  and  afterwards  sold 
it  so  as  to  lose  10  per  cent. :  IIOAV  much  did  he  get  for  it? 

Operation. 

$840  purchase  price.  Having  found  the  sum  lost,  (Art. 

.10  per  cent.  loss.  388,)  subtract  it  from  the  cost,  and 

$84.00  sum  lost.  the  remainder  is  obviously  the  sell- 

$756.00  selling  price.  ing  price.     Hence, 

445*  To  find  how  any  article  must  be  sold,  in  order  to  gain 
or  lose  a  given  rate  per  cent,  when  the  cost  is  given. 

First  find  the  amount  of  profit  or  loss  on  the  purchase  price  at 
the  given  rate,  as  in  the  last  Case  ;  then  the  amount  thus  found 
added  to,  or  subtracted  from  the.  purchase  price,  as  the  case  may  be, 
will  give  the  selling  price  required. 

12.  A  grocer  bought  a  quantity  of  cheese  for  $130.67:    for 
how  much  must  he  sell  it, 'to  gain  20  per  cent.  ? 

13.  Bought  a  stock  of  goods  for  $3460  :  for  how  much  must 
they  be  sold,  to  gain  22£  per  cent.  ? 

14.  Bought  a  quantity  of  flour  for  $5245  :  for  how  much  must 
it  be  sold,  to  gain  13  per  cent.  ? 

15.  Bought  2500  biles  of  cotton  for  $30575,  which  were  sold 
at  a  less  of  3£  per  cent. :  what  did  they  fetch  ? 

Q.UEST.— 445.  What  is  the  method  of  finding  how  any  article  must  be  sold,  in  order  to 
gain  or  lose  a  given  per  cent.  1  446.  How  is  the  rate  per  cent,  of  profit  or  loss  found,  when 
the  cost  and  selling  price  are  given  7 


286  PROFIT    AND    LOSS.  [SECT.   XII 

CASE    III. 

To  find  the  RATE  PER  CENT,  of  profit  or  loss,  the  cost  and  sell- 
ing price  being  given. 

16.  If  a  merchant  buys  a  quantity  of  butter  for  $75,  and  sells 
it  for  $90,  what  per  cent,  profit  will  he  make? 

Analysis. — Subtracting  the  cost  from  the  selling  price,  shows 
that  he  gained  $15.  Now  15  dollars  are  -ff-  of  75  dollars  ;  there- 
fore he  gained  ff  of  his  outlay,  or  the  purchase  price  of  the  goods. 
And  -ft  reduced  to  a  decimal,  is  equal  to  20  hundred ths,  or  20  per 
cent.  (Art.  387.  Obs.  3.) 

Or,  we  may  reason-  thus:  If  $75  (outlay)  gain  15  dollars,  $1 
will  gain  ^V  of  $15.  And  $15 -r- 75  =  .20,  the  same  as  before. 

446.  Hence,  to  find  the  rate  per  cent,  of  profit  or  loss,  when 
the  cost  and  selling  prices  are  given. 

First  find  the  amount  gained  or  lost  by  subtraction  ;  then  make 
the  gain  or  loss  tJie  numerator  and  the  purchase  price  the  denomina- 
tor of  a  common  fraction  ;  reduce  this  fraction  to  a  decimal,  and 
the  result  will  be  the  per  cent,  required.  (Art.  337.) 

Or,  simply  annex  ciphers  to  the  profit  or  loss,  and  divide  it  by 
the  cost  j  the  quotient  will  be  the  per  cent. 

OBS.  1.  As  per  cent,,  signifies  hundredths,  the  first  two  decimal  figures  which 
occupy  the  place  of  hundredths,  are  properly  the  per  cent.;  the  other  decimals 
are  -parts  of  1  per  cent.  After  obtaining  two  decimal  figures,  there  is  some- 
times an  advantage  in  placing  the  remainder  over  the  divisor,  and  annexing 
it  to  the  decimals  thus  obtained.  (Art.  387.  Obs.  3  ) 

2.  It  should  be  remembered  that  the  percentage  which  is  gained  or  last,  is 
always  calculated  on  the  purchase  price,  or  the  sum  paid  for  the  article,  ind 
not  on  the  selling  price,  or  sum  received,  as  it  is  often  supposed. 

17.  Bought  a  quantity  of  cotton  at  6£  cents  per  yard,  and  sold 
it  alt  8  cents :  what  per  cent,  was  the  profit  ? 

18.  Bought  a  quantity  of  calico,  at  12  cents  per  yard,  and  sold 
it  at  ]^£  cents :  what  per  cent,  was  the  profit? 

19.  Bought  a  lot  of  corn,  at-  45  cents  per  bushel,  and  sold  it  at 
88  cents  :  what  per  cent,  was  the  loss  ? 

QUKST,  -Obs.  What  figures  properly  signify  the  per  cent.  ?  What  do  the  other  decimal 
figures  on  the  lipbt  of  hundredths  denote  ?  On  what  is  the  per  cen  i.  gained  ot  lost  calc» 


ART.  446.]  PROFIT  AND  LOSS.  287 

20.  A  grocer  bought  a  pipe  of  wine  for  $252,  and  retailed  it 
at  12-£  cents  per  gill :  what  per  cent,  did  he  make? 

21.  A  man  bought  a  house  for  $4325,  and  sold  it  for  $5216: 
what  per  cent,  did  he  make  ? 

22.  A  speculator  invested  $75000  in  stocks,  which  he  sold  for 
877225 :  what  per  cent,  did  he  make  by  the  operation? 

CASE    IV. 

To  find  the  COST,  the  selling  price  and  per  cent,  gained  or  lost 
being  given. 

23.  A  man  sold  a  lot  of  salt  for  $360,  which  was  20  per  cent, 
more  than  cost :  what  did  he  pay  for  the  salt  ? 

Analysis. — The  cost  is  -HHi  of  itself,  and  the  gain  is  -iVV  of  the 
cost.  (Art.  386.)  Now  iU+YuQjr=ioi ;  hence,  the  selling  price 
is  -HHI-  of  the  cost.  The  question  then  is  this:  $360  is  -Hfr  of 
what  sum  ?  If  $360  is  -j-fg-  of  a  certain  sum,  -^  of  that  sum  is 
-rirr  of  $360.  Now  $360~120=$3,  and  -U$=$3X100,  which 
is  $300.  Ans. 

Or,  if  we  divide  $360,  the  selling  price,  by  the  fraction  -HHK 
the  quotient  $300,  will  be  the  cost.  (Art.  234.) 

PROOF. — $300X.20=$60.00  the  gain;  (Art.  388;) 
and  $300+$60=$360,  the  selling  price. 

24.  A  miller  sold  a  lot  of  flour  for  $170,  which  was  15  per 
cent,  less  than  cost :  how  much  did  the  flour  cost  him  ? 

Analysis. — Reasoning  as  before,  the  cost  is  -f{H|-  of  itself,  and 
the  loss  is  tW  of  the  cost.  Now  -HHJ — -iW^i'VV ;  consequently 
the  selling  price  is  -fifo  Oi  tne  cost.  The  question  therefore  ia 
this :  $170  is  -ffa  of  what  sum  ?  If  $170  is  T^JT  of  a  certain  sum, 
rfo-  is  -fa  of  $170.  Now  $!70-7-85=$2,  and  i££=$2xlOO, 
which  is  $200.  Ans. 

Or,  thus:  Since  lie  lost  15  per  cent.,  he  realized  only  85  cents 
on  $1  outlay.  Therefore,  if  85  cents,  selling  price,  requires  $1 
outlay,  $170,  selling  price,  will  require  as  many  dollars  outlay  aa 
85  cents  are  contained  times  in  $170 ;  and  $170^.85  =  $200. 

PROOF.— $200 X-15=$30.00,  the  loss  ;  (Art.  388  ;) 
and  $200 — $30=$170,  the  selling  price.     Hence, 


288  PROFIT    AND    LOSS.  [SECT.  Xll. 

447.  To  find  the  cost  when  the  selling  price  and  the  per  cent. 
gained  or  lost  are  given. 

Divide  the  selling  price  by  $1,  increased  or  diminished  by  the  per 
cent,  gained  or  lost,  as  the  case  may  be,  and  tJie  quotient  will  be  lh* 
cost  required. 

Or,  make  the  given  per  cent,  added  to  or  subtracted  from  100,  as 
tlte  case  may  be,  the  numerator,  and  100  the  denominator  of  a  com- 
mon fraction ;  then  divide  the  selling  price  by  this  fraction,  and 
tlie  quotient  will  be  the  cost. 

OBS.  1.  It  is  not  unfrequently  supposed  that  if  we  find  the  percentage  on 
the  selling  price  at  the  given  rate,  and  add  the  percentage  thus  found  to,  or 
subtract-  it  from,  the  selling  price,  as  the  case  may  be,  the  sum  or  remainder 
will  be  the  cost.  This  is  a  mistake,  and  leads  to  serious  errors  in  the  result. 
It  will  easily  be  avoided  by  remembering,  that  the  basis  on  which  profit  and 
loss  are  calculated,  is  always  the  purchase  price  or  sum  paid  for  the  articles 
under  consideration.  (Art.  446.  Obs.  2.) 

25.  A  grocer  sold  a  quantity  of  cheese  for  $530,  which  was  15 
per  cent,  more  than  cost :  what  was  the  cost  ? 

26.  A  man  sold  a  carriage  for  $175,  which  was  15  per  cent, 
less  than  cost :  what  was  the  cost  ? 

27.  A  man  sold  a  farm  for  $2360,  which  was  10  per  cent,  less 
than  cost :  what  did  he  give  for  it  ? 

28.  An  importer  sold  a  library  for  $3078,  which  was   12£  per 
cent,  advance  on  the  cost :  how  much  did  it  cost  him  ? 

29.  A  merchant  sold  a  cargo  of  crockery  for  $12000,  which  was 
8  per  cent,  less  than  cost :  what  was  the  cost  ? 

30.  A  commission  merchant  sold  a  lot  of  cloths  for  $7265,  which 
was  15  per  cent,  more  than  cost:  how  much  did  they  cost? 

31.  A  builder  sold  a  house  for  $17450,  which  was  2  per  cent. 
less  than  cost :  what  was  the  cost  ? 

32.  A  broker  sold  stocks  to  the  amount  of  $45000,  which  was 
5£  per  cent,  advance  :  what  was  the  cost  ? 

33.  A  manufacturer  sold  a  quantity  of  carpeting  for  $63240, 
which  was  50  per  cent,  more  than  the  cost  of  the  materials  :  what 
did  the  materials  cost  ? 

QUEST. — 147.  How  is  the  cost  found,  when  the  selling  price  and  the  rate  per  cent, 
gained  or  lost,  are  given?  Obs.  What  mistake  is  sometimes  made  in  finding  the  os'1 
How  may  it  be  avoided  1 


ARTS.  447-450.  |  DUTIES.  289 

DUTIES. 

448»  DUTIES,  in  commerce,  signify  a  sum  of  money  required 
by  Government  to  be  paid  on  imported  goods. 

OBS.  1.  In  every  port  of  entry  in  the  United  States,  the  Government  has  an 
establishment,  called  a  Custom  House,  at  which  the  duties  on  all  foreign  goods 
entered  at  that  port,  are  to  be  paid. 

2.  The  persons  appointed  to  inspect  the  cargoes  of  vessels  engaged  in  foreign 
commerce,  to  examine  the  invoices  of  goods,  collect  the  duties,  &c.,  are  called 
custom  louse  officers. 

449,  Duties  are  of  two  kinds,  specific  and  ad  valorem.  A  spe- 
cific duty  is  a  certain  sum  imposed  on  a  ton,  hundred  weight, 
hogshead,  gallon,  square  yard,  foot,  <fec.,  without  regard  to  the 
value  of  the  article. 

Ad  valorem  duties  are  those  which  are  imposed  on  goods,  at  a 
certain  per  cent,  on  their  value  or  purchase  price. 

Note. — The  term  ad  valorem  is  a  Latin  phrase,  signifying  according  to,  or 
upon  the  value. 

45O»  Before  specific  duties  are  imposed,  it  is  customary  to 
make  certain  deductions  called  tare,  draft  or  tret,  leakage,  &c. 

Tare  is  an  allowance  of  a  certain  number  of  pounds  made  for 
the  box,  cask,  &c.,  which  contains  the  article  under  consideration. 

Draft  or  Tret  is  an  allowance  of  a  certain  per  cent,  (usually  4 
per  cent.)  on  the  weight  of  goods  for  waste,  or  refuse  matter. 

Leakage  is  an  allowance  of  a  certain  per  cent,  (usually  2  per 
cent.)  for  the  waste  of  liquors  contained  in  casks,  &c. 

OBS.  1.  AH  duties,  both  specific  and  ad  valorem,  are  regulated  by  the  Gov- 
ernment, and  have  been  different  at  different  times  and  in  different  countries. 

2.  The  allowances  or  deductions  for  draft,  tare,  leakage,  &c.,  are  different  on 
different  articles,  and  are  also  regulated  by  law. 

3.  In  buying  and  selling  groceries  in  large  quantities,  allowances  are  some- 
times made  for  draft,  tare,  leakage,  &c.,  similar  to  those  in  reckoning  duties. 


QUEST.— 448.  What  are  duties  in  commerce?  449.  Of  how  many  kinds  ate  ..hey? 
What  arc  specific  duties  ?  Ad  valorem  duties  ?  Note.  What  is  the  meaning  of  the  term 
ad  valorem  ?  450.  What  deductions  are  made  before  specific  duties  are  imposed  1  What 
Is  tare?  Draft  or  tret?  Leakage?  Obs.  How  are  duties  regulated?  Are  allowance* 
for  draft,  &c.,  ever  made  in  buying  and  selling  groceries  ! 


290  DUTIES.  [SECT.  XII 

CASE  I. —  Calculation  of  Specific  Duties. 

Ex.  1.  What  is  the  specific  duty  on  15  hhds.  of  molasses,  at 
10  cents  per  gallon,  allowing  2  per  cent,  for  leakage? 

Analysis, — Since  there  are  63  gallons  in  one  hhd.,  in  15  hhds. 
there  are  15  times  as  many,  and  63  gals.  X  15  =  945  gals.  But  2  per 
cent,  of  945  gals,  is  equal  to  945 X. 02,  or  18.9  gals. ;  (Art.  388  ;) 
and  945  gals. — 18.9  gals.  =  926.1  gals.,  the  net  gallons.  Now  if 
he  duty  on  1  gallon  is  10  cents,  on  926.1  gals.it  is  926.1  X-10  — 
$92.61,  the  duty  required.  Hence, 

451.  To  find  the  specific  duty  on  any  given  merchandise. 

First  deduct  the  legal  draft,  tare,  leakage,  &c.,  from  the  given 
quantity  of  goods  ;  then  multiply  the  remainder  by  the  given  duty 
per  gallon,  pound,  yard,  &c.,  and  the  product  will  be  the  duty  re~ 


2.  If  the  specific  duty  on  tea  is  12  cents  a  pound,  how  much 
will  it  be  on  30  chests,  each  weighing  115  Ibs.,  allowing  12  Ibs. 
per  chest  for  draft  ? 

3.  At  4  cents  a  pound,  what  is  the  specific  duty  on  160  drums 
of  figs,  weighing  28  Ibs.  apiece,  allowing  2£  Ibs.  a  drum  for  tare? 

4.  At  15  cents  a  pound,  what  is  the  specific  duty  on  63  chests  of 
opium,  each  weighing  150-lbs.,  allowing  10  Ibs.  per  chest  for  draft? 

5.  At  3£  cents  a  pound,  what  is  the  specific  duty  on  250  bags 
of  coffee,  weighing  65  Ibs.  apiece,  allowing  4  per  cent,  for  tret  ? 

6.  What  is  the  specific  duty,  at  6  cents  a  pound,  on  173  kegs  of 
tobacco,  each  weighing  125  Ibs,,  allowing  6  Ibs.  per  keg  for  tare? 

7.  At  5i  cents  a  pound,  what  is  the  specific  duty  on  430  boxes 
of  paints,  weighing  175  Ibs.  a  box,  reckoning'  the  tare  at  15  Ibs. 
per  box  ? 

8.  At  8  cents  per  gallon,  what  is  the  specific  duty  on  140  Ibs. 
of  olive  oil,  allowing  2  per  cent,  for  leakage  ? 

9.  At  22  cents  per  gallon,  what  is  the  specific  duty  on  50  hhds. 
of  wine,  allowing  2  per  cent,  for  leakage  ? 

10.  At  7£  cents  per  pound,  what  is  the  duty  on  345  sacks  of 
almonds,  weighing  75  Ibs.  apiece,  allowing  3  per  cent,  for  tare  ? 

QUEST.— 451.  How  are  specific  duties  calculated" 


ARTS.  151-453.]  DUTIES.  291 

CASE  II. — Calculation  of  Ad  Valorem  Duties. 

452.  When  duties  are  imposed  upon  the  actual  cost  of  mer- 
chandise, there  are  of  course  no  deductions  to  be  made ;  conse- 
quently we  have  only  to  find  the  legal  per  cent,  on  the  amount  of 
the  given  invoice,  or  cost  of  the  goods,  and  it  will  be  the  duty 
required. 

Ex.  11.  What  is  the  ad  valorem  duty,  at  25  per  cent.,  on  a 
$ase  of  bombazines,  invoiced  at  $450  ? 

Solution. — $450 X-25=$l  12.50,  the  ad  valorem  duty.     Hence, 

453.  To  find  the  ad  valorem  duty  on  any  given  merchandise. 
Multiply  the  amount  of  the  given  invoice  by  the  legal  per  cent., 

and  the  product  will  be  the  duty  required.  (Art.  324.) 

OBS.  1.  An  invoice  is  a  written  statement  of  merchandise,  with  the  value  or 
prices  of  the  articles  annexed. 

2.  The  law  requires  that  the  invoice  shall  be  verified  by  the  owner,  or  one 
of  the  owners  of  the  goods,  certifying  that  the  invoice  annexed  contains  a  true 
and  fail  kf nl  account,  of  the  actual  costs,  thereof,  and  of  all  charges  thereon,  arftl 
no  other  different  discount,  bounty,  or  drawback,  but  such  as  has  been  actually 
allowed  on  the  same;  which  oath  shall  be  administered  by  a  consul,  or  com- 
mercial agent  of  the  United  States,  or  by  some  public  officer  duly  authorized 
to  administer  oaths  in  the  country  where  the  goods  were  purchased,  and  the 
same  shall  be  duly  certified  by  the  said  consul,  &c.  Fraud  on  the  part  of  the 
owners,  or  the  consul,  &c.,  who  administers  the  oath,  is  visited  with  a  heavy 
penalty. — Lajos  of  the  United  Stales. 

12.  What  is  the  ad  valorem  duty,  at  20  per  cent.,  on  an  invoice 
of  broadcloths  which  cost  $1240  in  Manchester? 

]  3.  What  is  the  ad  valorem  duty,  at  34  per  cent.,  on  an  invoice 
of  silks,  which  cost  $2110  in  Italy? 

14.  What  is  the  duty,  at  25  per  cent.,  on  a  quantity  of  indigo, 
the  invoice  of  which  is  $1968  ? 

15.  What  is  the  duty  on  a  bale  of  Irish  linens,  which  cost 
$3187,  at  33  per  cent.? 

16.  At  25  per  cent.,  what  is  the  duty  on  an  invoice  of  hosiery, 
amounting  to  $2863  ? 

QUEST — 453.  How  are  ad  valorem  duties  calculated  ?  Obs.  What  is  an  invoice?  What 
does  the  law  require  respecting  the  invoice  of  imported  goods  1 

•13* 


292  ASSESSMENT  [SECT.  XII. 

17.  At  33-£  per  cent.,  what  is  the  duty  on  an  invoice  of  mousse- 
line  de  laines,  amounting  to  $3690  ? 

18.  At  35  per  cent.,  what  is  the  duty  on  an  invoice  of  watches, 
amounting  to  $45385  ? 

19.  What  is  the  duty,  at  20  per  cent.,  on  an  invoice  of  boots 
and  shoes,  amounting  to  $63212  ? 

20.  What  is  the  duty,  at  15  per  cent.,  on  a  quantity  of  ready- 
made  clothing,  worth  $18714? 

21 .  What  is  the  duty  on  $37241  worth  of  spices,  at  30  per  ct.  ? 

22.  What  is  the  duty  on  $46210  worth  of  liquor,  at  37£  per 
cent.  ? 

23.  At  22  per  cent.,  what  is  the  duty  on  $71685  worth  of 
crockery  ? 

ASSESSMENT  OF   TAXES. 

454*  A  Tax  is  a  sum  of  money  assessed  on  individuals  for 
the  support  of  Government,  Corporations,  Parishes,  Districts,  <fcc. 
Taxes  levied  by  the  Government,  are  assessed  either  on  the  person 
or  property  of  the  citizens.  When  assessed  on  the  person,  they 
are  called  poll  taxes,  and  are  usually  a  specific  sum.  Those  as- 
sessed on  the  property  are  usually  apportioned  at  a  certain  per 
cent,  on  the  amount  of  real  estate  and  personal  property  of  each 
citizen  or  taxable  individual. 

OBS.  Property  is  divided  into  two  kinds,  viz :  real  estate  and  personal  prop- 
erty. The  former  denotes  possessions  that  are  fixed ;  as  houses,  lands,  &c. 
The  latter  comprehends  all  other  property ;  as  money,  stocks,  notes,  mortgages, 
ships,  furniture,  carriages,  cattle,  tools,  &c. 

455.  When  a  tax  of  any  given  amount  is  to  be  assessed,  the 
first  thing  to  be  done  is  to  obtain  an  inventory  of  the  amount  of 
taxable  property,  both  personal  and  real,  in  the  State,  County, 
Corporation,  or  District,  by  which  the  tax  is  to  be  paid ;  also  the 
amount  of  property  of  every  citizen  who  is  to  be  taxed,  together 
with  the  number  of  Polls. 

QUKST.— 454.  What  are  taxes  ?  Upon  what  are  they  assessed  ?  When  assessed  upon 
the  person,  what  are  they  called  ?  When  assessed  upon  the  property,  how  are  they  ap- 
portioned 1  Obs.  How  is  property  divided  ?  What  does  real  estate  denote  ?  What  I* 
personal  property  7  455.  When  a  tax  is  to  be  assessed,  wltat  is  the  first  step  ? 


ARTS.  454-456.]  OF  TAXES.  293 

OBS.  I.  By  the  number  of  polls  is  meant  the  number  of  taxable  individuals , 
which  usually  includes  every  native  or  naturalized  freeman  over  the  age  of  21, 
and  under  70  years.  In  Massachusetts  poll  taxes  are  assessot!  -jpon  every 
male  inhabitant  of  the  state,  between  the  ages  of  16  ajid  70  years,  whether  a 
citizen  or  an  alien.* 

2.  When  any  part  or  the  whole  of  a  tax  is  assessed  upon  the  polls,  each 
citizen  is  taxed  a  specific  sum,  without  regard  to  the  amount  of  property  he 
possesses. 

Ex.  1.  The  tax  assessed  by  a  certain  town  is  $990;  its  prop- 
erty both  personal  and  real,  is  valued  at  $28000,  and  it  contains 
300  polls,  which  are  assessed  50  cents  apiece.  What  per  cent,  is 
the  tax ;  that  is,  how  much  is  the  tax  on  a  dollar ;  and  how  much 
is  a  man's  tax  who  pays  for  3  polls,  and  whose  property  is  valued 
at  $1500?  * 

Solution. — Since  1  poll  pays  50  cents,  300  polls  must  pay  300 
times  50  cents,  which  is  $150.  Now  $990 — $150=$840,  the 
sum  to  be  assessed  on  the  property.  Now  if  $28000  is  to  pay 
$840,  $1  must  pay^rsforr  of  $840;  and  $840-r$28000=$.03, 
or  3  per  cent.  Finally,  the  tax  on  $1500,  the  amount  of  the 
man's  property,  at  3  percent.,  is  $1500X-03=$45 ;  and  $45  + 
$1.50  (3  polls)=$46.50,  the  man's  tax.  Hence, 

456.  To  assess  a  State,  County,  or  other  tax. 

I.  First  find  the  amount  of  tax  on  all  the  j>olls,  if  any,  at  the 
given  rate,  and  subtract  this  sum  from  the  whole  tax  to  be  assessed. 
Then  dividing  the  remainder  by  the  whole  amount  of  taxable  prop- 
erty in  the  State,  County,  d'c.,  the  quotient  will  be  the  per  cent,  or 
tax  on  one  dollar. 

II.  Multiply  the  amount  of  each  marts  property  by  the  tax  on 
one  dollar,  and  the  product  will  be  the  tax  on  Ids  property. 

III.  Add  each  mans  poll  tax  to 'the  tax  he  pays  on  his  property^ 
and  the  amount  will  be  his  whole  tax. 

PROCF. —  When  a  tax  bill  is  made  out,  add  together  the  taxes 
of  all  the  individuals  in  the  town,  district,  &c.,  and  if  the  amount 
is  equal  to  the  whole  tax  assessed,  the  work  is  right. 

CinicsT. — Obs.  What  is  meant  by  the  number  of  polls  ?  456.  How  are  taws  asseu*tf  * 
When  a  tax  bill  is  made  pat,  how  is  its  correctness  proved  ? 

*  Revised  Statutes  of  Massachusetts. 


294  ASSESSMENT  [SECT.    XII. 

2.  A  certain  corporation  is  taxed  $537.50  ;  the  whole  propeHy 
of  the  corporation  is  valued  at  $35000,  and  there  are  50  polls 
which  are  assessed  25  cents  apiece.     What  per  cent,  is  the  tax ; 
and  how  much  is  a  man's  tax,  who  pays  for  2  polls,  and  whose 
property  is  valued  at  $4240. 

Operation. 

Multiply  $.25  the  tax  on  1  poll, 
By     50  the  number  of  polls. 

$12.50  Amount  on  polls. 

But  $537.50 — 12.50=$525,  the  sum  assessed  on  the  corpora- 
tion ;  and  $525-r$35000=.015,  the  per  cent,  or  tax  on  $1. 
Now  $4240X.015=$63>60,  the  tax  on  the  man's  property, 
And  .25X2=       .50,  the  tax  for  polls. 

Am.  $64.10,  whole  tax. 

3.  What  is  B's  tax,  who  pays  for  3  polls,  and  whose  property  is 
valued  at  $3560  ? 

4.  What  is  C's  tax,  who  is  assessed  for  1  poll  and  $5350  ? 

5.  The  city  of  New  York  levied  a  tax  of  $1945600;  its  tax- 
able property  was  rated  at  $243200000 :  what  per  cent,  was  the 
tax? 

6.  What  was  A's  tax,  whose  property  was  valued  at  $10000  ? 

7.  What  was  B's  tax,  who  was  assessed  for  $15240  ? 

8.  What  was  C's  tax,  who  was  assessed  for  $35460  ? 

457.  Having  ascertained  the  expenditures  of  a  State,  County, 
Town,  &c.,  it  is  necessary  in  assessing  the  tax,  to  take  into  con 
sideration  the  expense  of  collecting  it.  Collectors  are  paid  a  certain 
per  cent,  commission  on  the  amount  collected  ;  (Art.  388.  Obs.  i  ;) 
consequently,  in  determining  the  exact  sum  to  be  assessed,  allow- 
ance must  be  made  not  only  for  the  commission  on  the  net  amount 
to  be  raised,  but  also  on  the  commission  itself;  for  the  commis- 
ion  is  to  be  paid  out  of  the  money  collected. 

9.  If  the  expenses  of  a  town  are  $950,  what  sum  must  be 
assessed  to  raise  this  amount,  with  5  per  cent,  commission  for 
collecting  it  ? 


ARTS.  i57-459.]  OF  TAXES.  295 

Analysis. — Since  the  commission  is  5  per  cent,  the  net  value 
of  $1  assessment  is  95  cents.  Therefore,  if  95  cents  net,  require 
$1  assessment,  8950  net,  will  require  as  many  dollars  assessment, 
as  95  cents  are  contained  times  in  $950;  and  $950-f-$.95=1000. 

Ans.  $1000. 

PROOF. — $1000X-05=$50,  the  commission; 

and  $1000 — $50=8950,  the  net  sum  required.     Hence, 

458.  -To  find  what  sum  must  be  assessed,  to  raise  a  given  net 
amount. 

Subtract  the  given  per  cent,  commission  from  $1,  and  the  re~ 
mainder  will  be  the  net  value  of  $1  assessment. 

Divide  the  net  amount  to  be  raised  by  the  neb  value  of  $1  assess  • 
ment,  and  the  quotient  will  be  the  sum  to  be  assessed. 

OBS.  To  meet  the  expense  of  collecting  a  tax,  assessors  not  unfrequently  cal- 
cujate  the  commission  at  the  given  per  cent,  on  the  net  amount  to  be  raised, 
and  add  it  to  the  tax  bill.  This  method  is  wrong,  and  leads  to  erroneous  re- 
sults. Thus,  on  a  tax  of  $1000,  at  5  per  cent,  commission,  the  net  amount  i 
$2.50  too  small ;  on  $100000.  the  error  is  $250 ;  on  $1000000,  it  is  $"2500. 

10.  What  sum  must  be  assessed  to  raise  a  net  amount  of  $8500., 
with  4  per  cent,  commission  for  collection  ? 

11.  What  sum  must  be  assessed  to  raise  $15400  net,  allowing 
4£  per  cent,  commission  for  collection  ? 

12.  Allowing  5   per  cent,  for  collection,  what  sum  must  be 
assessed  to  raise  $16475  net? 

13.  Allowing  3£  per  cent,  for  collection,  what  sum  must  h>€ 
assessed  to  raise  $32860  net? 

FORMATION    OF    TAX    BILLS. 

459.  In  making  out  a  tax  bill  for  a  Town,  District,  &c.,  hav- 
ing found  the  tax  on  $1,  it  is  advisable  to  make  a  table,  show- 
ing the  amount  of  tax  on  any  number  of  dollars  from  1  to  $10 ; 
then  from  $10  to  $100 ;  and  from  $100  to  $1000. 

14.  A  township  composed  of  16  citizens,  levies  a  tax  of  $5700  ; 
the  tywn  contains  30  polls,  which  are  assessed  50  cents  each,  and 

QUEST. — 458.  How  find  what  sum  must  be  assessed  to  raise  a  tax  of  a  given  amount? 


290 


ASSESSMENT 


[SECT.  XII. 


its  taxable  property  is  inventoried  at  $199500.  What  amount  of 
tax  must  be  raised  to  pay  the  debt  and  5  per  cent,  commission 
for  collection ;  and  what  is  the  tax  on  a  dollar  ? 

Solution. — The  sum  to  be  raised  is  $6000  ;  (Art.  458  ;)  and  the 
tax  is  3  cents  on  a  dollar.  (Art.  456.)     Now,  since  the  tax  on  $1 
is  $.03,  it  is  obvious  that  multiplying  $.03  by  2  will  be  the  tax  on 
$2  ;  multiplying  it  by  3,  will  be  the  tax  on  $3,  &c.,  as  seen  in  th 
following 

TABLE. 


$1  pays  $.03 

$10  pay  $.30 

$100  pay  $3.00 

2  "   .06 

20  "   .60 

200  "   6.00 

3  "   .09 

30  "   .90 

300  "   9.00 

4  "   .12 

40  "  1.20 

400  "  12.00 

5  "   .15 

50  "  1.50 

500  "  15.00 

6  "   .18 

60  "  1.80 

600  "  18.00 

7  "   .21 

70  "  2.10 

700  "  21.00 

8  "   .24 

80  "  2.40 

800  "  24.00 

9  "   .27 

90  "  2.70 

900  "  27.00 

10  "   .30 

100  "  3.00 

1000  "  30.00 

15.  In  the  above  assessment,  what  is  A.  B.'s  tax,  who  is  rated 
at  $2256,  and  pays  for  3  polls  ? 

Operation. 


$2000  pay  $60.00 

200     "        6.00 

50     "         1.50 

6     "  .18 

3  polls   "         1.50 

Amount,     $69.18 


$2256  =  2000  +  200+50  +  6  dollars. 
Now  if  we  add  together  the  tax  paid 
on  each  of  these  sums,  as  found  in  the 
table  above,  the  amount  will  be  the  tax 
on  $2256. 
A.  B.'s  tax  therefore  is  $69.18. 


16.  What  is  G.  A.'s  tax,  who  is  assessed  for  2  polls,  and  $2400 

17.  What  is  H.  B.'s  tax,  who  is  assessed  for  1  poll,  and  $3850  ? 

18.  What  is  W.  C.'s  tax,  who  is  assessed  for  3  polls  and  $1 5000? 

19.  E.  D.  is  assessed  for  $16024,  and  1  poll:  what  is  his  tax? 

20.  J.  F.  is  assessed  for  $10450,  and  2  polls:  what  is  his  tax? 

21.  T.  G.  is  assessed  for  $20680,  and  3  polls  :   what  is  his  tax? 

22.  W.  H.  is  assessed  for  $17530,  and  1  poll :  what  is  his  tax  ? 


ART.  4GO.]  OF  TAXES.  &"* 

23.  L.  J.  is  assessed  for  $8760,  and  1  poll :  what  is  his  tax? 

24.  W.  L.  is  assessed  for  $21000,  and  2  polls  :  what  is  his  tax? 

25.  J.  K.  is  assessed  for  $6530,  and  2  polls:  what  is  his  tax? 

26.  G.  L.  is  assessed  for  $13480,  and  1  poll :  what  is  his  tax  ? 

27.  F.  M.  is  assessed  for  $12300,  and  3  polls  :  what  is  his  tax? 
26.  C.  P.  is  assessed  for  $15240,  and  2  polls  :  what  is  his  tax': 

29,  J.  S.  is  assessed  for  $16000,  and  1  poll :  what  is  his  tax? 

30.  R.  W.  is  assessed  for  $18000,  and  2  polls  :  what  is  his  tax? 

Note. — Rate  Dills  for  schools  are  generally  apportioned  according  to  the 
number  of  days  each  scholar  has  attended.  Hence, 

46 O»  To  make  out  Rate  Bills  for  schools. 

First  find  the,  number  of  days  attendance  of  all  the  scholars, 
and  the  'whole  amount  of  expenses,  including  teacher's  salary,  fuel, 
repairs,  &c.  From  the  amount  of  expenses  deduct  the  public 
money,  if  any,  then  divide  the  remainder  by  the  whole  number  of 
days  attendance,  and  the  quotient  will  be  the  rate  per  day.  Finally, 
multiply  the  rate  per  day  by  the  number  of  days  attendance  of  each 
mans  children,  and  the  'product  will  be  his  tax. 

OPS.  In  New  York  and  some  other  states,  the  general  principle  is  to  include 
only  the  Teacher's  Salary  in  the  Rate  Bill.  (Revised  Statutes.  N.  Y.) 

?l.  A  certain  district  paid  $130  for  teacher's  salary,  $34  for 
bocvd,  $19.42  for  fuel,  and  $2.58  for  repairs ;  the  district  drew 
$30  public  money,  and  the  whole  number  of  days  attendance  was 
2400 :  what  was  the  rate  per  day ;  and  how  much  was  A's  tax, 
wh<-»  sent  115  days? 

F'vlution. — Amount  of  expenses,  $186 — $30=$156  ;  and  $156 
•7-2*00=4.065,  the  rate  per  day.  Now  $.065 X  115=$7.475, 
A's  tax  is  therefore  $7.475. 

3l\  If  the  expenses  of  a  district  are  $313.20,  and  the  whole 
attendance  3915  days,  what  is  B's  tax,  who  sends  167  days? 

33.  A   district  paid  their  teacher  $115,  and  their  fuel   cost 
£21  10  ;  it  drew  $38.50  public  money,  and  the  number  of  days 
attendance  was  1954 :  what  was  C's  tax,  who  sent  69  days? 

34.  The  expenses  of  a  district  were  $215.20,  and  the  number 
of  days  attendance  2150  :  what  was  D's  tax,  who  ser.t  134  days? 


298  ANALYSIS.  [SECT.  Xlil 

SECTION     XIII. 
ANALYSI.S. 

ART.  46 1  •  The  term  Analysis,  in  physical  science,  signifies  the 
resolving  of  a  compound  body  into  its  elements,  or  component  parts. 

ANALYSIS,  in  arithmetic,  signifies  the  resolving  of  numbers  into 
£he  factors  of  which  they  are  composed,  and  the  tracing  of  the 
relations  which  they  bear  to  each  other.  (Art.  95.  Obs.  2.) 

OBS.  In  the  preceding  sections  the  studen.t  has  become  acquainted  with  the 
method  of  analyzing  particular  examples  and  combinations  of  numbers,  and 
thence  deducing  general  'principles  and  rules.  But  analysis  may  be  applied 
with  advantage  not  only  to  the  development  of  mathematical  truths,  but  also  to 
the  solution  of  a  great  variety  of  problems,  both  in  arithmetic  and  practical 
life.  Indeed,  it  is  the  method  by  which  business  men  generally  solve  prac- 
tical questions.  A  little  practice  will  give  the  student  great  facility  in  it* 
application. 

462*  No  specific  directions  can  be  given  for  solving  examples 
by  analysis.  None  in  fact  are  requisite.  The  judgment,  from 
the  conditions  of  the  question,  will  suggest  tlie  process.  Hence, 
Analysis  may,  with  propriety,  be  called  the  COMMON  SENSE  RULE. 

OBS.  In  solving  questions  analytically,  it  may  be  remarked  in  general,  that 
we  reason  from  the  given  number  to  1,  then  from  1  to  the  number  required. 

Ex.  1.  If  60  yards  of  cloth  cost  $240,  what  will  85  yards  cost? 

Analytic  solution. — Since  60  yds.  cost  $240,  1  yd.  will  cost  -6-V 
of  $240 ;  and  -gV  of  $240  is  $4.  Now  if  1  yd.  costs  $4,  85  yds. 
will  cost  85  times  as  much;  and  $4X85=$340.  Ans. 

Or,  we  may  reason  thus :  85  yds.  are  -f-fr  of  60  yds. ;  therefore 
85  yds.  will  cost  ££  of  $240,  (the  cost  of  60  yds.)  and  f-g-  of  $240 
is  $240XfS=$340,  the  same  as  before.  (Arts.  210,  212.) 

OBS.  1.  Other  solutions  of  this  example  might  be  given;  but  our  preseat  c!>- 
jsct  is  to  show  how  this  and  similar  questions  may  be  solved  by  analysis.  The 

QUEST. — 461.  What  is  meant  by  analysis  in  physical  science?    What  in  arithmetic  1 
To  what  may  analysis  be  advantageously  applied  ?    462.  Can  any  particular  rules  be  pre-> 
scribed  for  solving  questions  by  analysis  ?     How  then  will  you  know  how  t  >  proceed 
Obs.  What  is  the  operation  of  solving  questions  by  analysis  called  7 


ARTS.  461,  462.]  ANALYSIS.  29!) 

former  method  is  the  simplest  and  most  strictly  analytic,  though  not  so  short 
as  the  latter.  It  contains  two  steps : 

First,  we  separate  the  given  price  of  GO  yds.  ($240)  into  60  equal  parts,  to 
find  the  value  of  one  part,  or  the  cost  of  1  yd.,  which  is  $4. 

Second,  we  multiply  the  price  of  1  yd.  ($4)  by  85,  the  number  of  yds.  whose 
cost  is  required,  and  the  product  is  the  answer  sought. 

2.  This  and  similar  questions  are  usually  placed  under  the  rule  of  Simple 
Proportion,  or  the  Ride  of  Three. 

3.  The  operation  of  solving  a  question  by  analysis,  is  called  an  analytic 
oluiion.  -  In  reciting  the  following  examples,  each  one  should  be  analyzed, 
i)d  the  reason  for  every  stej  given  in  full. 

2.  A  man  bought  a  horse,  and  paid  $45  down,  which  was  $•  of 
the  price  of  it :  what  did  he  give  for  the  horse  ? 

Analysis. — Since  $45  is  -f-  of  the  price,  the  question  resolves 
itself  into  this :  $45  is  f  of  what  sum  ?  If  $45  is  f  of  a  certain 
sum,  -f  is  £  of  $45  ;  and  £  of  $45  is  $9.  Now  if  $9  is  1  seventh, 
7  sevenths  are  7  times  as  much;  and  $9xV=$63.  Ans.  $63. 

PROOF. — $  of  $6 3 =$9,  and  5  sevenths  are  5  times  as  much, 
which  is  $45,  the  sum  he  paid  down  for  the  horse. 

Note. — In  solving  examples  of  this  kind,  the  learner  is  often  perplexed  in 
finding  the  value  of  -f ,  &c-  This  difficulty  arises  from  supposing  that  if  •&• 
of  a  certain  number  is  45,  \  of  it  must  be  \  of  45.  This  mistake  will  be 
easily  avoided  by  substituting  in  his  mind  the  word  parts  for  the  given  de- 
nominator. Thus,  if  5  parts  cost  $45,  \  part  will  cost  £  of  $45,  which  is  $9. 
But  this  part  is  a  seventh.  Now  if  1  seventh  cost  $9,  then  7  sevenths  will 
cost  7  times  as  much. 

3.  If  40  cords  of  wood  cost  $120,  how  much  will  100  cords 
sost? 

.  4.  Bought  35  tons  of  hay  for  $700:  how  much  will  16  tens 
cost? 

5.  What  cost  37  gallons  of  molasses,  at  $21  a  hogshead? 

6.  What  cost  1500  pounds  of  hay,  at  $14  per  ton? 

7.  What  cost  18  quarts  of  chestnuts,  at  $3  a  bushel? 

8.  If  55  tons  of  hemp  cost  $660,  what  will  220  tons  cost  at  the 
same  rate  ? 

9.  If  165  bushels  of  apples  cost  $132,  how  much  will  31  bushelg 
cost? 


300  ANALYSIS.  [SECT.  XIII. 

10.  It  72  bushels  of  peanuts  cost  $253.44,  what  will  a  pint 
cost  at  tne  same  rate  ? 

11.  If  150  acres  of  land  cost  $7000,  what  will  a  square  rod 
cost? 

12.  If  2  pipes  of  wine  cost  $315,  what  is  that  per  gill? 

13.  A  farmer  bought  a  yoke  of  oxen,  and  paid  $40  in  work, 
whbh  was  -f-  of  the  cost :  what  did  they  cost  ? 

14.  Bought  a  house,  and  paid  $030  in  goods,  which  was  -fo  of 
the  price  of  it :  what  was  the  cost  of  the  house  ? 

15  A  young  man  lost  $256  by  gambling,  which  was  -fg  of  all 
he  was  worth :  how  much  was  he  worth  ? 

16.  A  man  having  $1500,  paid  f  of  it  for  112£  acres  of  land: 
how  much  did  his  land  cost  per  acre  ? 

17.  If  a  stack  of  .hay  will  keep  350  sheep  90  days,  how  long 
will  it  keep  525  sheep  ? 

18.  If  440  bbls.  of  flour  will  last  15  men  55  months,  how  long 
will  the  same  quantity  last  28  men? 

19.  If  136  men  can  build  a  block  of  stores  in  120  days,  how 
long  will  it  take  15  men  to  build  it? 

20.  If  •£  of  a  pound  of  tea  cost  40  cents,  what  will  -£  of  a  pound 
cost  ? 

21.  If  -f  of  a  yard  of  broadcloth  cost  $2.50,  how  much  will  $ 
of  a  yard  cost  ? 

22.  Bought  -ft  of  a  ton  of  hay  for  $3.42  :  how  much  will  -ft  of 
a  ton  cost  ? 

23.  Bought  £$  of  a  hogshead  of  molasses  for  $38.19:  how 
much  will  ^o  of  a  hogshead  cost  ? 

24.  If  f  of  an  acre  of  land  cost  $108,  how  much  will  J-  of  an 
acre  cost  ? 

25.  If  f  of  a  barrel  of  beef  cost  $6.48,  how  much  will  f  of  a 
barrel  cost  ? 

26.  Paid  $4200  for  f  of  a  sloop :  how  much'  c?an  I  afford  to 
iell  -^  of  the  sloop  for  ? 

2*.  Sold  18£  baskets  of  peaches  for  $34 :  how  much  would  65-J- 
baskets  come  to  ? 

28.  If  I  pay  $60.50  for  building  20 Jr  rods  of  wall,  how  much 
must  I  pay  for  <215|  rods? 


AKT.  463.J  ANALYSIS.  301 

29.  A  man  can  hoe  a  field  of  corn  in  6  days,  and  a  boy  can  hoe 
H  in  9  days  :  how  long  will  it  take  them  both  together  to  hoe  it  ? 

Analysis. — Since  the  man  can  hoe  the  field  in  6  days,  in  1  day 
he  can  hoe  £  of  it ;  and  since  the  boy  can  hoe  it  in  9  days,  in  1 
day  he  can  hoe  \  of  it ;  consequently  in  1  day  they  can  both  hoe 
1 4.  1=-^  of  the  field.  (Art.  202.)  Now  if  -fa  of  the  field  requires 
them  both  1  day,  -fa  of  it  will  require  them  £  of  a  day,  and  \\ 
full  require  them  18  times  as  long,  or  -^  of  a  day,  which  is  equal 
to  3f  days.  Ans. 

30.  If  A  can  chop  a  cord  of  wood  in  4  hours,  and  B  in  G  hours, 
how  long  will  it  take  them  both  to  chop  a  cord  ? 

31.  A  can  dig  a  cellar  in  G  days,  B  in  9  days,  and  C  in  12 
days :  how  long  will  it  take  all  of  them  together  to  dig  it  ? 

32.  A  man  bought  25  pounds  of  tea  at  6s.  a  pound,  and  paid 
for  it  in  corn  at  4s.  a  bushel :  how  many  bushels  did  it  take  ? 

Analysis. — -If  1  Ib.  of  tea  costs  6s.,  25  Ibs.  will  cost  25  times 
as  much,  which  is  150s.  Again,  if  4s.  will  buy  1  bushel  of  corn, 
150s.  will  buy  as  many  bushels  as  4s.  is  contained  times  in  150s. ; 
and  150s.-f-4=37£  times.  Ans.  37i  bushels. 

463.  The  last  and  similar  examples  are  frequently  arranged 
under  the  rule  of  Barter. 

Barter  signifies  an  exchange  of  articles  of  commerce  at  prices 
agreed  upon  by  the  parties. 

OBS.  Such  examples  are  so  easily  solved  by  Analysis  that  a  specific  ride  fct 
them  is  unnecessary. 

33.  A  farmer  bought  110  Ibs.  of  sugar  at  18  cents  a  pound, 
and  paid  for  it  in  lard  at  5£  cents  a  pound :  how  much  lard  did 
it  take  ? 

34.  How  much  butter,  at  12^  cents  a  pound,  must  be  given  for 
250  Ibs.  of  tea,  at  75  cents  a  pound  ? 

35.  How  many  cords  of  wood,  at  $2£  per  cord,  must  be  given 
for  56  yds.  of  cloth,  at  &4-J-  per  yard  ? 

36.  How  many  pair  of  boots,  at  $4.50  a  pair,  must  be  given 
for  50  tons  of  coal  at  $9  per  ton  ? 


302  ANALYSIS.  [SECT.  XIII. 

37.  A,  B,  and  C,  united  in  business ;  A  pit  in  $250  ;  B,  $270  ; 
and  C,  $340 ;  they  gained  $258  :  what  was  each  man's  share  of 
the  gain  ? 

Analysis. — The  whole  sum  invested  is  $250-r-$270+$340  = 
$860.  Now  since  $860  gain  $258,  it  is  plain  $1  will  gain  -g^  of 
$258,  which  is  30  cents.  And 

If  $1  gains  30  cts.  $250  will  gain  $250X-30=$75,  A's  share, 
"  $1  "  "  $270  "  $270 X. 30=  81,  B's  share, 
"  $1  "  "  $340  "  $340X-30  =  102,  C's  share. 

Or,  we  may  reason  thus :   Since  the  sum  invested  is  $860, 
A's  part  of  the  investment  is  equal  to  •§£$,  or  -ff- ; 
B's  "  «  «  ^ffcortt; 

C's  "  "  "  £££,  or  if.  -  Consequently, 

A  must  receive  |f  of  the  whole  gain  $2 5  8 =$7 5  ; 
B  "H         "  "  258=   81; 

C  "if         "  "  258  =  102; 

PROOF. — The  whole  gain  is  $258.  (Ax.  11.) 

464.  When  two  or  more  individuals  associate  themselves  to- 
gether for  the  purpose  of  carrying  on  a  joint  business,  the  union 
is  called  a  partnership  or  copartnership. 

OBS.  The  process  by  which  examples  like  the  last  one  are  solved,  is  often 
called  Fellowship. 

38.  A  and  B  join  in  a  speculation ;  A  advances  $1500  and  B 
$2500  ;  they  gain  $1200  :  what  was  each  one's  share  of  the  gain  ? 

39.  A,  B,  and  C,  entered  into  partnership ;  A  furnished  $3000, 
B  $4000,  and  C  $5000;  they  lost  $1800:  what  was  each  one's 
share  of  the  loss  ? 

40.  A's  stock  is  $4200  ;  B's  $3600 ;  and  C's  $5400 ;  the  whole 
gain  is  $2400  :  what  is  the  gain  of  each  ? 

41.  A's  stock  is  $7560;    B's  $8240;    C's  $9300;   and  D's 
$6200 ;  the  whole  gain  is  $625  :  what  is  the  share  of  each  ? 

42.  A  bankrupt  owes  one  of  his  creditors  $400  ;  another  $500  \ 
and  a  third  $600;  his  property  amounts  to  $1000:  how  much 
can  he  pay  on  a  dollar ;  and  how  much  will  each  of  his  creditors 
receive  ? 

OBS.  The  solution  of  this  example  is  the  same  in  principle  as  that  of  Ex.  37 


ARTS.  464-4G6.]  ANALYSIS.  303 

46 5 •  Examples  like  the  preceding  are  commonly  arranged 
under  tlw  rule  of  Bankruptcy. 

Note..  —A  bankrupt,  is  a  person  who  is  insolvent,  or  unable  to  pay  his  just 
debts. 

43.  A  bankrupt  owes  $5000,  and  his  property  is  worth  $3500 : 
how  much  can  he  pay  on  a  dollar  ? 

4  t.  A  man  di(d  owing  $16400,  and  his  effects  were  sold  foi 
f-4  i  0(  :  what  per  cent,  did  his  estate  pay  ? 

45.  If  a  man  owes  A  $6240,  B  $8760,  and  C  $9000,  and  has 
but  $lloOO,  how  much  will  each  creditor  receive? 

46.  If  I  owe  $48000,  and  have  property  to  the  amount  of 
$32000,  what  per  cent,  can  I  pay  ? 

47  What  per  cent,  can  a  man  pay,  whose  liabilities  are 
$120000,  and  whose  assets  are  $45000? 

48.  What    per   cent,    can   a   man    pay,   whose   liabilities   are 
$1500000,  and  whose  assets  are  $150000? 

4G6.  It  often  happens  in  storms  and  other  casualties  at  sea, 
that  masters  of  vessels  are  obliged  to  throw  portions  of  their 
cargo  overboard,  or  sacrifice  the  ship  and  their  crew.  In  such 
cases,  the  law  requires  that  the  loss  shall  be  divided  among  the 
owners  of  the  vessel  and  cargo,  in  proportion  to  the  amount  of 
each  one's  property  at  stake. 

The  process  of  finding  each  man's  loss,  in  such  instances,  is 
called  General  Average. 

OBS.  The  operation  is  the  same  as  that  in  solving  questions  in  bankruptcy 
and  partnership. 

49.  A,  B  and  C,  freighted  a  ship  from  New  York  to  Liverpool ; 
A  had  on  board  100  tons  of  iron,  B  200  tons,  and  C  300  tons, 
in  a  storm  240  tons  were  thrown  overboard :  what  was  the  loss 
of  each  ? 

50.  A  packet  worth  $36000  was  loaded  with  a  cargo  valued  at 
$65000.     In  a  tempest  the  master  threw  overboard  $25250  worth 
of  goods  ;  what  per  cent,  was  the  general  average  ? 

51.  A  steam  ship  being  in  distress,  the  master  threw  \  of 
the  cargo  overboard ;    finding  she  still  labored,   he  afterwards 
threw  overboard  ^  of  what  remained.     The  steamer  was  worth 


304  ANALYSIS.  [SECT.  XIII. 

$120000,  and  the  cargo  $240000  :  what  per  cent,  was  the  general 
average,  and  what  would  be  a  man's  loss  who  owned  -J-  of  tl?.?  ship 
and  cargo? 

52.  A  man  mixed  25  bushels  of  peas  worth  6s.  a  bushel,  with 
15  bushels  of  corn  worth  4s.  a  bushel,  and  20  bushels  of  oats 
worth  3s.  a  bushel :  what  was  the  mixtuie  worth  per  bushel .' 

Analysis. — 25  bu.  peas  at  6s.  =  150s.,  value  of  the  peas; 
15  bu.  corn  at  4s.  =   60s.,         "       "       corn; 
and  20  bu.  oats  at  3s.  =   60s.,         "       "       oats. 
The  mixture=60  bu.  and  270s.,  value  of  whole  mixture. 

Now  if  60  bu.  mixture  are  worth  270s.,  1  bu.  mixture  is  worth 
jV  of  270s. ;  and  270s.-H60^4i-s.  Ans. 

PROOF. — 60  bu.  at  4£s.=270s.,  the  value  of  the  whole  mixture. 

467.  The  process  of  finding  the  value  of  a  compound  or  mix- 
tare  of  articles  of  different  values,  or  of  forming  a  compound 
which  shall  have  a  given  value,  is  called  Alligation.  Alligation  is 
usually  divided  into  two  kinds,  Medial  and  Alternate. 

OBS.  1.  When  the  prices  of  the  several  articles  and  the  number  or  quantity 
of  each  are  given,  the  process  of  finding  the  value  of  the  mixture,  as  in  the  last 
example,  is  called  Alligation  Medial. 

2.  When  the  price  of  the  mixture  is  given,  together  with  the  price  of  each 
article,  the  process  of  finding  how  much  6T  the  several  articles  must  be  taken 
to  form  the  required  mixture,  is  called  Alligation  Alternate  Alligation  Alter- 
nate embraces  three  varieties  of  examples,  which  are  pointed  out  in  the  follow- 
ing notes. 

53.  If  you  mix  40  gallons  of  sperm  oil  worth  8s.  per  gallon, 
with  60  gallons  of  whale  oil  worth  3s.  per  gallon,  what  will  the 
mixture  be  worth  per  gallon  ? 

54.  At  what  price  per  pound  can  a  grocer  afford  to  sell  a  mix- 
ture of  30  Ibs.  of  tea  worth  4s.  a  pound,  and  40  Ibs.  worth  7s.  a 
pound  ? 

55.  If  120  Ibs.  of  butter  at  10  cts.  a  pound  are  mixed  with  24 
Ibs.  at  8  cts.  and  24  Ibs.  at  5  cts.  a  pound,  what  is  the  mixture 
worth  ? 

56.  A  tobacconist  had  three  kinds  of  tobacco,  worth  15,  18, 
and  25  cents  a  pound:    what  is  a  mixture  of  100  Ibs   of  each 
worth  per  pound  ? 


ART  467.]  ANALYSIS/  305 

57.  A  liquor  dealer  mixed  200  gallons  of  alcohol  worth  50  cts. 
a  gallon,  with  100  gallons  of  brandy  worth  $1.75  a  gallon:  what 
was  the  value  of  the  mixture  per  gallon  ? 

58.  A  grocer  sells  imperial  tea  at  10s.  a  pound,  and  hyson  at 
4s. :  what  part  of  each  must  he  take  to  form  a  mixture  which  he 
can  affc  rd  to  sell  at  6s.  a  pound  ? 

Note  \ .—  It  will  be  observed  in  this  example  that  the  price  of  the  mixture 
W*J.  ilsc  the  price  of  the  several  ar ticks  or  ingredients  are  given,  o  find  what 
]~w  jf  each  the  mixture  must  contain. 

Analysis. — Since  the  imperial  is  worth  10s.  and  the  required 
mixture  6s.,  it  is  plain  he  would  lose  4s.  on  every  pound  of  impe- 
rial which  he  puts  in.  And  since  the  hyson  is  worth  4s  .a  pound 
and  the  mixture  6s.,  he  would  gain  2s.  on  every  pound  of  hyson 
he  puts  in.  The  question  then  is  this :  How  much  hyson  must 
he  put  in  to,  make  up  for  the  loss  on  1  Ib.  of  imperial  ?  If  2s. 
profit  require  1  Ib.  of  hyson,  4s.  profit  will  require  twice  as  much, 
or  2  Ibs.  He  must  therefore  put  in  2  Ibs.  of  hyson  to  1  Ib.  of  im- 
perial. 

PROOF — 2  Ibs.  of  hyson,  at  4s.  a  pound,  are  worth  8s.,  and 
1  Ib.  of  imperial  is  worth  10s.  Now  8s.+10s.=18s.  And  if 
3  Ibs.  mixture  are  worth  18s.,  1  Ib.  is  worth  |  of  18s.,  which  is 
6s.,  the  price  of  the  mixture  required. 

59.  A  farmer  has  oats  which  are  worth  20  cts.  a  bushel,  rye 
.55  cts.,  and  barley  60  cts.,  of  which  he  wishes  to  make  a  mixture 
worth  50  cts.  per  bushel :    what  part  of  each  must  the  mixture 
contain  ? 

Analysis. — The  prices  of  the  rye  and  barley  must  each  be  com- 
pared with  the  price  of  the  oats.  If  1  bu.  oats  gains  30  cts.  in 
the  mixture,  it  will  take  as  many  bu.  of  rye  to  balance  it,  as  5  cts. 
(the  loss  per  bu.)  are  contained  times  in  30  cts.,  viz :  6  bu.  Again, 
since  1  bu.  oats  gains  30  cts.,  it  will  take  as  many  bushels  of  bar- 
Jey  to  balance  it,  as  10  cts.  (the  loss  per  bu.)  are  contained  times 
in  30  cts.,  viz:  3  bu.  Hence,  the  mixture  must  contain  2  paita 
of  oats,  6  parts  rye,  and  3  parts  barley. 

60.  Tf  a  man  have  four  kinds  of  sugar  worth,  8,  9,  11,  and  12 


306  ANALYSIS.  [SECT.  XIII. 

cents  a  pound  respectively,  how  much  of  each  kind  must  he  take 
to  form  a  mixture  worth  10  cents  a  pound  ? 

NoLe  2. — In  examples  like  the  preceding,  we  compare  two  kinds  together, 
one  of  a  higher  and  the  other  of  a  lower  price  than  the  required  mixture ;  then 
compare  the  other  two  kinds  in  the  same  manner.  In  selecting  the  pairs  to 
be  compared  together,  it  is  necessary  that  the  price  of  one  article  shall  be 
above,  and  the  other  below  the  price  of  the  mixture.  Hence,  when  there  are 
aeveral  articles  to  be  mixed,  some  cheaper  and  others  dearer  than  the  mixture, 
a  variety  of  answers  may  be  obtained.  Thus,  if  we  compare  the  highest  and 
lowest,  then  the  other  two,  the  mixture  will  contain  1  part  at  8  cts. ;  1  part  at 

0  cts. ;  1  part  at  11  cts.;  and  1  part  at  12  cts.     Again,  by  comparing  those 
at  8  and  1 1  cts.,  and  those  at  9  and  12  cts.  together,  we  obtain  for  the  mixture 

1  part  at  8  els. ;  2  parts  at  1 1  cts. ;  2  parts  at  9  cts. ;  and  1  part  at  12  cts. 
Other  answers  may  be  found  by  comparing  the  first  with  the  third  and 

fourth ;  and  the  second  with  the  fourth,  &c. 

61.  A  goldsmith  having  gold  16,  18,  23,  and  24  carats  fine, 
wished  to  make  a  mixture  21  carats  fine  :  what  part^of  each  must 
the  mixture  contain  ? 

62.  A  farmer  had  30  bu.  of  corn  worth  6s.  a  bu.,  which  he 
wished  to  mix  with  oats  worth  3s.  a  bu.,  so  that  the  mixture  may 
be  worth  4s.  per  bu. :  how  many  bushels  of  oats  must  he  use  ? 

Note  3. — In  this  example,  it  will  be  perceived,  that  the  price  of  the  mix- 
ture, with  the  prices  of  the  several  articles  and  the  quantity  of  one  of  them 
are  given,  to  find  how  much  of  the  other  article  the  mixture  must  contain. 

Analysis. — Reasoning  as  above,  we  find  that  the  mixture  (with  • 
out  regard  to  the  specified  quantity  of  corn)  in  order  to  be  worth 
4s.  per  bu.,  must  contain  2  bu.  of  oats  to  1  bu.  of  corn.  Hence, 
if  1  bu.  of  corn  requires  2  bu.  of  oats  to  make  a  mixture  of  the 
required  value,  30  bu.  of  corn  will  require  30  times  as  much ;  and 

2  bu.X  30  =  60  bu.,  the  quantity  of  oats  required. 

63.  A  merchant  wished  to  mix  100  gallons  of  oil  worth  80  cts. 
per  gallon,  with  two  other  kinds  worth  30  cts.  and  40  cts.  per  gal- 
lon, so  that  the  mixture  may  be  worth  60  cts.  per  gallon :  how 
many  gallons  of  each  must  it  contain  ? 

64.  A  merchant   has    Havana   coffee  at  12  cts.  and  Java  at 
18  cts.  per  pound,  of  which  he  wishes  to  make  a  mixture  of  150 
Ibs.,  which  he  can  sell  at  16  cts.  a  pound:  how  much  of  each 
must  he  use  ? 


A.RT.  468.]  ANALYSIS.  307 

Note  4. — In  this  example,  the  whole  quantity  to  be  mixed,  the  jnce  of  the 
aixture,  and  the  prices  of  the  several  articles  are  given,  to  fi.id  Juw  much  of 
«ach  must  be  taken. 

Analysis. — On  1  Ib.  of  the  Havana  it  is  obvious  he  will  gain 
4  cts.,  and  on  1  Ib  of  the  Java  he  will  lose  2  cts. ;  therefore  to 
balance  the  4  cts.  gain  he  must  put  in  2  Ibs.  of  Java;  that  is,  the 
mixture  must  contain  1  part  of  Havana  to  2  parts  of  Java.  Now 
if  3  Ibs.  mixture  require  1  Ib.  Havana,  150  Ibs.  mixture,  (the  quan- 
tity required,)  will  require  as  many  pounds  of  Havana  as  3  is  con- 
tained times  in  150,  viz  :  50  Ibs.  But  the  mixture  contains  twice 
as  much  Java  as  Havana,  and  50  Ibs  X  2  =  100  Ibs. 

Ans.  50  Ibs.  Havana,  and  100  Ibs.  Java. 

65.  It  is  required  to  mix  240  Ibs.  of  different  kinds  of  raisins, 
worth  8d.,  12d.,  18d.,  and  22d.  a  pound,  so  that  the  mixture  may 
be  worth  lOd.  a  pound  :  how  much  of  each  must  be  taken  ? 

66.  If  10  horses  consume  720  quarts  of  oats  in  6  days,  how 
long  will  it  take  30  horses  to  consume  1*728  quarts? 

Analysis. — Since  10  horses  will  consume  720  qts.  in  6  days, 
1  horse  will  consume  -fa  of  720  qts.  in  the  same  time ;  and  -fa  of 
720  qts.  is  72  qts.  And  if  1  horse  will  consume  72  qts.  in  6  days, 
in  1  day  he  will  consume  •£•  of  72  qts.,  which  is  12  qts.  Again, 
if  12  qts.  last  1  horse  1  day,  1728  qts.  will  last  him  as  many 
days  as  12  qts.  are  contained  times  in  1728  qts.,  viz:  144  days. 
Now  if  1  horse  will  consume  1728  qts.  in  144  days,  30  horses 
will  consume  them  in  -fa  of  the  time;  and  144  d.^-30=4f. 

Ans.  30  horses  will  consume  1728  qts.  in  4f  days. 

468.  This  and  similar  examples  are  usually  placed  under  the 
rule  of  Compound  Proportion,  or  Double  Rule  of  Three. 

67.  If  15  horses  consume  40  tons  of  hay  in  30  weeks,  how  many- 
horses  will  it  require  to  consume  56  tons  in  70  weeks  ? 

68.  If  8  men  can  make  9  rods  of  wall  in  12  days,  how  long  wil 
\\  take  10  men  to  make  36  rods? 

69.  If  35  bbls.  of  water  will  last  950  men  7  months,  how  many 
men  will  1464  bbls.  of  water  last  1  month? 

T.H.  14 


308  ANALYSIS.  LSEC 

70.  If  13908  men  consume  732  bbis.  of  flour  in  2  months,  in 
how  long  time  will  425  men  consume  175  bbls.  ? 

71.  If  the  interest  of  $30  for  12  months  is  $2.10,  how  much  ia 
the  interest  of  $1560  for  6  months? 

72.  If  the  interest  of  $750  for  8  months  is  $28,  how  much  is 
the  interest  of  $16425  for  6  months  ? 

73.  A  man  being  asked  how  much  money  he  had,  replied  that 
•f,  •£-,  and  f  of  it  made  $980  :  what  amount  did  he  have  ? 

Analysis. — It  is  plain  that  -f  +  f -f -f=f£.  (Art.  202.)  The 
question  then  resolves  itself  into  this  :  $980  are  ff  of  what  sum  ? 
Now  if  $980  are  ff-  of  a  certafn  sum,  -gV  is  -fc  of  $980  ;  and  $980 
•i-49=$20,  and  -ft  is  $20X24=$480.  Ans. 

PROOF. — |  of  $480=$320  ;  ?  of  $480=$360  ;  and  i  of  $480 
=$300.  Now  $320+$360+$300=$980,  according  to  the  con- 
ditions of  the  question. 

469.  This  and  similar  examples  are  placed  under  the  rule  of 
Position.  The  shortest  and  easiest  method  of  solving  them  is  by 
Analysis. 

74.  A  sailor  having  spent  •$•  of  his  money  for  his  outfit,  depos- 
ited -f  of  it  in  a  savings  bank,  and  had  $50  left:  how  much  had 
he  at  first  ? 

75.  A  man  laid  out  -^  of  his  money  for  a  house,  i  for  furniture, 
and  had  $1500  left;  how  much  had  he  at  first? 

76.  A  man  lost  %  °f  his  money  in  gambling,  -J-  in  betting,  and  | 
spent  -f  in  drinking ;  he  had  $259  left :  how  much  had  he  at  first? 

77.  What  number  is  that  -f  and  £  of  which  is  102  ? 

78.  What  number  is  that  i,  |,  i,  and  |  of  which  is  450  ? 

79.  What  number  is  that  -£•  and  £  of  which  being  added  to 
itself,  the  sum  will  be  164  ? 

80.  What  number  is  that  •£  of  which  exceeds  f-  of  it  by  18  ? 

81.  A  post  stands  40  feet  above  water,  -J-  in  the  water,  and  J- 
in  the  ground  :  what  is  the  length  of  the  post  ? 

82.  What  will  376  yds.  of  muslin  cost,  at  2s.  and  6d.  per  yd.? 

Analysis. — 2s.  6d.=£i-.  Now  if  1  yd,  costs  £-£,  376  yds.  will 
cost  376  times  as  much;  and  £iX376=£47.  Ans. 


ARTS.  469,  470. 


ANALYSIS. 


309 


83.  If  1  yard  of  silk  costs  50  cents,  what  will  256  yards 
cost? 

Analysis. — 50  cts.=$£.  Now  if  1  yd.  costs  $i,  256  yds.  will 
cost  256  times  as  much;  and  $iX256=$128.  Ans. 

47O.  Examples  like  the  preceding,  in  which  the  price  of  a 
single  article  is  an  aliquot  part  of  a  dollar,  &c.,  are  usually  classed 
under  the  rule  of  Practice. 

Practice  is  denned  by  a  late  English  author  to  be  "  an  abridged 
method  of  performing  operations  in  the  rule  of  proportion  by  means 
of  aliquot  parts  j  and  it  is  chiefly  employed  in  computing  the 
prices  of  commodities." 

OBS.  After  giving  several  tables  of  aliquot  parts  in  money,  weight,  and 
measure,  the  same  author  proceeds  to  divide  his  subject  into  twelve  subdivi- 
sions or  cases,  and  gives  a  specific  ruk  for  each  case,  to  be  committed  to  mem- 
ory by  the  pupil.  It  is  believed,  however,  that  so  many  specific  rules  are  worse 
than  useless.  They  have  a  tendency  to  prevent  the  exercise  of  thought  and 
reason,  while  they  tax  the  time  and  memory  of  the  student  with  a  multiplicity 
of  particular  directions  for  the  solution  of  a  class  of  examples,  which  his  com- 
mon sense,  if  permitted  to  be  exercised,  will  solve  more  expeditiously  by 
Analysis. 

TABLE    OF    ALIQUOT    PARTS    OF    $1,    £l,    AND    Is. 


Parts  of  a  Dollar. 


Parts  of  a  Pound  sterling. 


Parts  of  a  Shilling  sterling. 


50  cts.=$i 
33|  cts.=*j 

20  cts.— -^ 


10    cts- 


5     cts.—  %- 


10s.   = 
6s.  8d.= 

5s.  = 
4s.  = 
3s.  4d.= 
2s.  6d.= 
2s.  = 
Is.  8d.= 
Is.  = 


=£    shil. 


6  pence 
4  pence  = 
3  pence  = 
2  pence : 
l£  pence : 
1  penny: 
£  penny: 

7  pence : 

8  pence:  .   .  ,  . 

9  pence  =£s.+-£s. 


shil. 
=i  shil. 
=i  shil. 
=i  shil. 
=T"*  shil. 
=^~f  shil. 


Note. — If  the  price  itself  is  not  an  aliquot  part  of  SI,  or  £1,  &c.,  it  may 
sotaetimes  be  divided  into  such  parts  as  will  be  aliquot  parts  of  $1,  £1,  &c., 
or  which  vrill  be  aliquot  parts  of  each  other.  Thus,  87£  cts.  is  not  an  a^quot 
part  of  $1,  but  87£  cts.  =50+254-12$  cts.  Now  50cts.=$J ;  25cts.=$| ;  and 
12$  cts.=$J.  Or  thus  :  50  cte.  =-.$$,  25  cts.=£$  of  50  cts.,  and  12$  cts.=:$  of 
25  cts. 


310  ANALYSIS.  [SttCT.   XIII. 

84.  What  will  680  bu.  of  wheat  cost,  at  87 £  cts.  per  bushel? 
Analysis. — It  is  plain,  if  the  price  were  $1  per  bu.,  the  cost  of 

680  bu.  would  be  $680.     Hence, 

Were  the  price  50  cts.  the  cost  would  be  £  of  $680,  which  is  $340 
25  cts.          "  "  i-of  $680,  which  is  $170 

12£  cts.       "  "          -i-  of  $680,  which  is  $  85 

But  since  the  price  is  50-f-25-hl2£  cents,  the  cost  must  be  $595 
'Or,  thus:  $1X680=$680,  the  cost  at  $1  per  bushel. 
At  50  cts.,  or  $£,  it  will  be  £  of  $680,  or  $340 

"    25  cts.,  i  of  50  cts.,     "     "     i  of  $340,  or  $170 
"    12i  cts.,  -^  of  25  cts.,    "     "     i  of  $170,  or  $  85 
Therefore  the  whole  cost  is  $595.  Ans. 

85.  What  cost  478  yards  of  cashmere,  at  50  cts.  per  yard? 

86.  What  cost  1560  Ibs.  of  tea,  at  75  cts.  per  pound? 

87.  What  cost  2400  gals,  of  molasses,  at  37£  cts.  per  gal.  ? 

88.  What  cost  1800' yds.  of  satinet,  at  62£  cts.  per  yard? 

89.  At  25  cts.  per  bushel,  what  cost  1470  bu.  of  oats? 

90.  At  33^  cts.  a  pound,  what  cost  1326  Ibs.  of  ginger? 

91.  At  6|  cts.  per  roll,  what  cost  3216  rolls  of  tape? 

92.  At  8|  cts.  per  pound,  what  cost  4200  Ibs.  of  lard  ? 

93  At  12£  cts.  per  dozen,  what  cost  1920  doz.  of  eggs? 

94  At  16f  cts.  a  pound,  what  cost  4524  Ibs.  of  figs  ? 

9f     At  66-f  cts.  per  yard,  what  cost  1620  yds.  of  sarcenet? 

9f     What  cost  840  bu.  of  rye,  at  $f  per  bushel  ? 

9V    What  cost  690  yds.  of  cloth,  at  6s.  8d.  per  yard  ? 

A  vtlysis.^At  £1  per  yard  the  cost  would  be  £690.  But 
6s,.  Vd.  is  £^-;  therefore  the  cost  must  be  £  of  £690,  which  is 
£2?<\  Ans. 

98.  What  cost  360  gals,  of  wine,  at  16s.  per  gallon? 
Analysis.— 16s.  =  10s.  +  5s.-f Is.      Now   10s.  =  £i;    5s.=£|; 
U.=i  of  5s. 

1  f  the  price  were  £l  per  gal.,  the  cost  of  360  gals,  would  be  £360. 
At  10s.,  £i,       it  will  be  i  of  £360,  or  £180 
"    5s.,  i  of  I Os.,       "       i  of  £180,  or  £  90 
"    Is.,  i  of  5s.,         "       i  of  £  00,  or  £  18 
Therefore  the  whole  cost  is  £288.  Ans. 


ART.  471." 


ANALYSIS. 


311 


99.  What  cost  1240  yds.  of  flannel,  at  3s.  4d.  per  yard? 

100.  What  cost  2128  Ibs.  of  spice,  at  2s.  6d.  ptr  pound? 

101.  What  cost  5250  yds.  of  lace,  at  6d.  per  yard  ? 
102    What  cost  56480  yds.  of  tape,  at  l|d.  per  yard? 

471.  Notwithstanding  the  law  requires  accounts  to  be  kept 
in  Federal  Money,  goods  are  frequently  sold  at  prices  stated  in 
the  denominations  of  the  old  state  currencies. 

When  the  price  per  yard,  pound,  &c.,  stated  in  those  currencies, 
is  an  aliquot  part  of  a  dollar,  the  answer  may  be  easily  obtained 
in  Federal  Money. 

TABLE  OF  ALIQUOT  PARTS  IN  DIFFERENT  STATE  CURRENCIES. 


Parts  of  a  Dollar, 
New  York  Currency. 

Parts  of  a  Dollar, 
New  K/igland  Currency. 

*   Parts  of  a  Shilling, 
N.  E.  and  N.  Y.  Currency. 

4  shil.  =$£ 

3  shil.  =$£ 

6    pence  =%    shil. 

2s.  8d.=$i 

2  shil.  —  $^- 

4    pence  =•$•    shil. 

2  shil.  =$| 

Is.  6d.=t4 

3    pence  =•}    shil. 

Is.  4d.=$£ 

1  shil.  =$i 

2    pence  =£    shil. 

1  shil.  =$1 

4s.=$i-j-$£ 

1-^-  pence  =i-    shil. 

5s.=4i+$i 

5s.  =  $i+$i 

1    penny  =-fV  shil. 

•  Note. — 1 .  In  N.  Y.  currency  8s.  make  SI ;  in  N.  E.  currency  6s.  make  $1. 
From  example  103  to  119  inclusive,  the  prices  are  given  in  N.  Y.  currency; 
from  example  120  to  132  inclusive,  they  are  given  in  N.  E.  currency.  For 
the  mode  of  reducing  the  different  State  currencies  to  each  other  and  to  Federal 
Money,  see  Section  XVII. 

103.  At  Is.  4d.  per  yard,  what  cost  726  yds.  of  cambric  ? 
Analysis. — If  the  J3rice  were  $1  per  yard,  the  cost  would  be 

$1X726=4726.  But  Is.  4d.=$i-;  therefore  the  cost  must  be 
i  of  $726,  which  is  $121.  Am. 

104.  What  cost  896  bu.  of  wheat  at  6s.  per  bushel? 
Analysis. — 6s.=4s.+2s.     Now  4s.— $£;  and  2s.=£  of  4s. 
At  $1  a  bushel  the  cost  would  be  1896. 

At  4s.,  $i,         it  will  be  £  of  $896,  or  $448 
"    2s,,  i  of  4s.,     "     "    i  of  $448,  or  $224 
Therefore  the  whole  cost  is  $672.  Ans. 

Or,  thus :  6s=$| ;  therefore  the  number  of  bu.  minus  |  of  itself 
will  be  the  cost,  ind  896—224  (|  of  896)=672.      Ans.  $672. 


312  ANALYSIS.  [SECT.  XIII. 

105.  What  cost  752  yds.  of  balzorine,  at  2s.  8d.  per  yard? 

106.  What  cost  1232  yds.  of  calico,  at  Is.  6d.  per  yard? 

107.  What  cost  763  Ibs.  of  pepper,  at  Is.  3d.  a  pound? 

108.  What  cost  1116  bu.  of  apples,  at  Is.  4d.  per  bushel? 

109.  What  cost  1920  yds.  of  shirting,  at  Is.  2d.  per  yard? 

110.  At  6s.  a  basket,  what  will  1560  baskets  of  peaches  cost? 

111.  At  5s.  4d.  a  pound,  what  will  1200  Ibs.  of  tea  come  to? 

Note.— 2.  Since  5s.  4<1.  is  \  less  than  SI,  it  is  plain  1200—  400=$800.  Ans 

112.  At  7s.  per  yard,  what  will  432  yds.  of  crape  cost? 

113.  At  6s.  8d.  a  pound,  what  cost  972  Ibs.  of  nutmegs? 

114.  At  2s.  8d.  a  pair,  what  cost  864  pair  of  cotton  hose? 

115.  At  1-^d.  a  yard,  how  much  will  2800  yds.  of  tape  come  to  ? 

116.  What  cost  16^8  yds.  of  flannel,  at  4s.  per  yard? 

117.  What  cost  2560  bu.  of  oats,  at '2s.  per  bushel? 

118.  What  cost  9600  Ibs.  of  wool,  at  2s.  6d.  a  pound  ? 

119.  What  cost  3200  Ibs.  of  sugar,  at  6d.  per  pound? 

120.  What  cost  600  yds.  of  damask,  at  5s.,  N.  E.  cur.,  per  yard  ? 
Note.— 3.  5s.  N.  E.  cur.  is  |  less  than  SI ;  hence,  GOO— 100=S500.  Ans 

121.  What  cost  2500  bu.  of  potatoes,  at  Is.  6d.  per  bushel? 

122.  What  cost  1440  yds.  of  gingham,  at  2s.  per  yard?  ' 

123.  How  much  will  4848  chickens  cost,  at  Is.  apiece? 

124.  How  much  will  1680  slates  cost,  at  Is.  6d.  apiece? 

125.  How  much  will  920  turkeys  cost,  at  4s.  6d.  apiece? 

126.  What  cost  4860  Ibs.  of  butter,  at  Is.  Id.  per  pound? 

127.  What  cost  1260  melons,  at  8d.  apiece? 

128.  What  cost  2340  Ibs.  of  tea,  at  4s.  a  pound? 

129.  What  cost  240  bu.  of  peas,  at  4s.  6d.  per  bushel? 

130.  What  cost  720  pair  of  gloves,  at  5s.  3d.  a  pair? 

131.  What  cost  360  bushels  of  corn,  at  3s.  per  bushel? 

132.  What  cost  7686  Ibs.  of  butter,  at  Is.  per  pound? 
133    What  cost  960  yds.  of  silk,  at  5s.  per  yard? 

134.  What  will  75  Ibs.  of  butter  cost,  at  $16.80  per  cwt.  ? 

135.  What  will  125  Ibs.  of  wool  cost,  at  $36  per  hundred  ? 

136.  What  will  15  ewt.  of  hemp  cost  at  $60  per  ton  ? 

137.  What  will  2500  Ibs.  of  iron  cost,  at  $72  per  ton? 

138.  What  cost  H  acre  of  land,  at  $120  per  acre  ? 


ARTS.  472-475.J  RATIO.  313 

SECTION    XIV. 
RATIO  AND   PROPORTION 

ART.  472*  In  comparing  numbers  or  quantities  with  each 
other,  we  may  inquire,  either  how  much  greater  one  of  the  num 
bers  or  quantities  is  than  the  other  ;  or  how  many  times  one  of 
them  contains  the  other.  In  finding  the  answer  to  either  of  these 
inquiries,  we  discover  what  is  called  the  relation  between  the  two 
numbers  or  quantities. 

473.-  The  relation  between  the  two  quantities  thus  compared, 
is  of  two  kinds  : 

First,  that  which  is  expressed  by  their  difference. 

Second,  that  which  is  expressed  by  the  quotient  of  the  one  di- 
vided by  the  other. 

47  4.  RATIO  is  that  relation  between  two  numbers  or  quanti- 
.    ties,  which  is  expressed  by  the  quotient  of  the  one  divided  by  the 
other.     Thus,  the  ratio  of  6  to  2  is  6-^-2,  or  3  ;  for  3  is  the  quo- 
tient of  6  divided  by  2. 

OBS.  The  relation  between  two  numbers  or  quantities  denoted  by  their  dif- 
ference, is  sometimes  called  arithmetical  ratio  ;  while  that  denoted  by  the  quo- 
tient of  the  one  divided  by  the  other,  is  called  geometrical  ratio.  Thus  4  is  the 
arithmetical  ratio  of  8  to  4  ;  and  2  is  the  geometrical  ratio  of  8  to  4. 

But  as  the.  term  arithmetical  ratio  is  merely  a  substitute  for  the  word  differ- 
ence, the  term  di  (Terence,  in  the  succeeding  pages,  is  used  in  its  stead  ;  and  when 
the  word  ratio  simply  is  used,  it  signifies  that  which  is  denoted  by  the  qiwiient 
of  the  one  divided  by  the  other,  as  in  the  article  above, 

475.  The  two  given  numbers  thus  compared,  when  sptken 
of  together,  are  called  a  couplet  ;  when  spoken  of  separately,  they 
are  called  the  terms  of  the  ratio. 

The  first  term  is  the  antecedent  ;  and  the  last,  the  consequent. 


72.  In  how  many  ways  are  numbers  or  quantities  compared?  474.  What  Is 
ratio?  475.  What  are  the  two  given  numbers  called  when  spoken  of  together?  When 
•poke*  of  separately  ? 


314  RATIO.  [SECT.  ^ 

47  6»  Ratio  is  expressed  in  two  ways : 

First,  in  the  form  of  a  fraction,  making  the  antecedent  the 
numerator,  and  the  consequent  the  denominator.  Thus,  the  ratio 
of  8  to  4  is  written  -f ;  the  ratio  of  12  to  3,  V>  &c. 

Second,  by  placing  two  points  or  a  colon  ( : )  between  the  num- 
bers compared.  Thus,  the  ratio  of  8  to  4  is  written  8:4;  the 
ratio  of  12  to  3,  is  12  :  3,  &c.  The  expressions  -f-,  and  8  :  4,  are 
of  the  same  import,  and  one  may  be  exchanged  for  the  other,  at 
pleasure. 

OBS  1.  The  sign  ( :  )  used  to  denote  ratio ,  is  derived  from  the  sign  of  divi- 
sion, -F- )  the  horizontal  line  being  omitted.  The  Engfish  mathematicians 
put  the  antecedent  for  the  numerator,  and  the  consequent  for  the  denomina- 
tor as  above ;  but  the  French  put  the  consequent  for  the  numerator  and  the 
antecedent  for  the  denominator.  The  English  method  appears  to  be  equally 
simple,  while  it  is  confessedly  the  most  in  accordance  with  reason. 

2.  In  order  that  concrete  numbers  may  have  a  ratio  to  each  other,  they  must 
necessarily  express  objects  so  far  of  the  same  nature,  that  one  can  be  properly 
said  to  be  equal  to,  or  greater,  or  less  than  the  other.  (Art.  294.)  Thus  a  foot 
has  a  ratio  to  a  yard ;  for  one  is  three  times  as  long  as  the  other ;  but  a  foot 
has  not  properly  a  ratio  to  an  hour,  for  one  cannot  be  said  to  be  longer  or 
shorter  than  the  other. 

477.  A  direct  ratio  is  that  which  arises  from  dividing  the 
antecedent  by  the  consequent;  as  6-:- 2.  (Art.  474.) 

478.  An  inverse,  or  reciprocal  ratio,  is  the  ratio  of  the  recip- 
rocals of  two  numbers.  (Art.  160.  Def.  10.)      Thus,  the  direct 
ratio  of  9  to  3,  is  9  :  3,  or  f ;  the  reciprocal  ratio  is  \ :  ^,  or  -J-f- 
•|  =  f ;  (Art.  229;)  that  is,  the  consequent  3,  is  divided  by  the 
antecedent  9. 

Note. — The  term  inverse,  signifies  inverted.     Hence, 

An  inverse,  or  reciprocal  ratio  is  expressed  by  inverting  the  frac- 
tion which  expresses  the  direct  ratio ;  or  when  the  notation  is  by 
points,  by  inverting  the  order  of  the  terms.  Thus,  8  is  to  4,  in- 
versely, as  4  to  8. 

QUEST. — 476.  In  how  many  ways  is  ratio  expressed  ?  The  first?  The  second?  Oft*. 
W7hich  of  the  terms  do  the  English  mathematicians  put  for  the  numerator  ?  Which  do 
the  French  ?  In  order  that  concrete  numbers  may  have  a  ratio  to  each  other,  what  kind 
of  objects  must  they  express  ?  477.  What  is  a  direct  ratio  ?  478.  What  is  an  inverse  at 
reciprocal  ratio  ?  How  is  a  reciprocal  ratio  expressed  by  a  fraction  ?  How  ty  points  ? 


ARTS.  476-481.]  RATIO.  315 

479*  A  simple  ratio  is  a  ratio  which  has  but  one  antecedent 
and  one  consequent,  and  may  be  either  direct  or  inverse  ;  as  9  :  3, 


48O»  A  compound  ratio  is  the  ratio  of  the  products  of  th<* 
corresponding  terms  of  two  or  more  simple  ratios.  Thus, 

The  simple  ratio  of  9  :     3  is  3  ; 

•4  And      "       "      of  8  :     4  is  2  ; 

The  ratio  compounded  of  these  is  72  :  12  =  6. 

OBS   1.  A  compound  ratio  is  of  the  same  nature  as  any  other  ratio.     Th» 
term  is  used  to  denote  the  origin  of  the  ratio  in  particular  cases. 
2.  The  compound  ratio  is  equal  to  the  product  of  the  simple  ratios. 

Ex.  1.  What  is  the  ratio  of  27  to  9  ?  Ans.  3. 

2.  What  is  the  ratio  of  8  to  32  ?  Ans.  -J-. 

Required  the  ratio  of  the  following  numbers  : 

3.  14  to     7.         13.  324  to    81.  23.  63  Ibs.  to  9  oz. 

4.  36  to    9.         14.  802  to    99.  24.  68  yds.  to  17  yds. 

5.  54  to    6.         15.        9  to    45.  25.  40  yds.  to  20  ft. 

6.  108  to  18.  16.  17  to    68.  26.  60  miles  to  4  fur. 

7.  144  to  24.  17.  13  to    52.  27.  45  bu.  to  3  pks. 

8.  156  to  17.  18.  27  to  135.  '  28.  6  gals,  to  1  hhd. 

9.  261  to  29.  19.  53  to  212.  29.  3  qts.  to  20  gal. 

10.  567  to  63.         20.     47  to  329.  30.  £l  to  15s. 

11.  '405  to  45.         21.  18  Ibs.  to  6  Ibs.         31.  15s.  to  £3. 

12.  576  to  64.         22.  28  Ibs.  to  4  Ibs.         32.  £10  to  lOd. 

48  1.  From  the  definition  of  ratio  and  the  mode  of  expressing 
it  in  the  form  of  a  fraction,  it  is  obvious  that  the  ratio  of  two  num- 
bers is  the  same  as  the  value  of  a  fraction  whose  numerator  and 
denominator  are  respectively  equal  to  the  antecedent  and  conse- 
quent of  the  given  couplet  ;  for,  each  is  the  quotient  of  the  numer- 
ator divided  by  the  denominator.  (Arts.  474,  185.) 

OBS.  From  the  principles  of  fractions  already  established,  we  may,  there- 
fore, deduce  the  following  truths  respecting  ratios. 

QUEST.—  479.  What  is  a  simple  ratio  1  480.  What  is  a  compound  ratio  ?  Obs  Doei  it 
differ  i/1  its  nature  from  other  ratios  1  What  is  the  term  used  to  denote  ? 

14* 


316  RATIO.  [SECT.  XIV. 

482*  To  multiply  ike  antecedent  of  a  couplet  by  any  number, 
multiplies  tlie  ratio  by  that  number  ;  and  to  divide  the  antecedent, 
divides  the  ratio :  for,  multiplying  the  numerator,  multiplies  the 
value  of  the  fraction  by  that  number,  and  dividing  the  numerator, 
divides  the  value.  (Arts.  186,  187.) 

Thus,  the  ratio  of         16  :  4  is  4  ; 

The  ratio  of  16  X  2  :  4  is  8,  which  equals  4X2; 

And      "  16-f-2  :  4  is  2,  which  equals  4-7-2. 

OBS.  With  a  given  consequent  the  greater  the  antecedent,,  the  greater  the 
ratio ;  and  on  the  other  hand,  the  greater  the  ratio,  the  greater  the  antece- 
dent. (Art.  187.  Obs.) 

48  3«  To  multiply  the  consequent  of  a  couplet  by  any  number, 
divides  the  ratio  by  that  number  y  and  to  divide  the  consequent, 
multiplies  the  ratio:  for,  multiplying  the  denominator,  divides  the 
value  of  the  fraction  by  that  number,  and  dividing  the  denomina- 
tor, multiplies  the  value.  (Arts.  188,  189.) 

Thus,  the  ratio  of  16  :  4        is  4  ; 

The  "  16  :  4X2  is  2,  which  equals  4-^2 ; 

And         "  16  :  4-i-2  is  8,  which  equals  4X2. 

OBS.  With  a  given  antecedent,  the  greater  the  consequent,  the  less  the  ratio ; 
and  the  greater  the  ratio,  the  less  the  consequent.  (Art.  189.  Obs.) 

484»  To  multiply  or  divide  both  the  antecedent  and  consequent 
of  a  couplet  by  the  same  number,  does  not  alter  the- ratio  :  for,  mul- 
tiplying or  dividing  both  the  numerator  and  denominator  by  the 
same  number,  does  not  alter  the  value  of  the  fraction.  (Art.  191.) 

Thus,  the  ratio  of        12:4        is  3  ; 

The  "  12X2:  4X2  is  3; 

And         "  12-7-2:  4-;- 2  is  3. 

485*  If  the  two  numbers  compared  are  equal,  the  ratic  is 
unit  or  1,  and  is  called  a  ratio  of  equality.     Thus,  the  ratio  of 
6X2  :  12  is  1  ;  for  the  value  of  H=  1.  (Art.  196.) 


QUKST. — 482.  What  is  the  effect  of  multiplying  the  antecedent  of  a  couplet  by  any  num 
her?  Of  dividing  the  antecedent  ?  483.  What  is  the  effect  of  multiplying  the  consequent 
by  any  number  ?  Of  dividing  the  consequent  1  Why  1  484.  What  is  the  effect  of  mul- 
tiplying or  dividing  both  the  antecedent  and  consequent  by  the  same  number?  Why? 
485.  When  the  two  numbers  compared  are  equal,  what  is  the  ratio  7  What  is  it  called  ? 


ARTS.  482-488.]  RATIO.  317 

486.  If  the  antecedent  of  a  couplet  is  greater  than  the  conse- 
quent, the  ratio  is  greater  than  a  unit,  and  is  called  a  ratio  of 
greater  inequality.     Thus,  the  ratio  of  12  :  4  is  3  ;  for  the  value 
of  Y=3.  (Art.  196.) 

487.  If  the  antecedent  is  less  than  the  consequent,  the  ratio 
is  less  than  a  unit,  and  is  called  a  ratio  of  less  inequality.     Thus, 
(he  ratio  of  3  :  G  is  £ ,  or  £ ;  for  £ =$.  (Art.  195.) 

OBS.  1.  The  direct  ratio  of  two  fractions  which  have  a  common  numerator, 
i*  the  same  as  the  reciprocal  ratio  of  their  denominators.  Thus,  the  ratio  of 
^  :  -f  is  the  same  as  -J- ;  •£-,  or  8 :  4. 

2.  The  ratio  of  two  fractions  which  have  a  common  denominator,  is  the 
same  as  the  ratio  of  their  numerators.  Thus,  the  ratio  of  |- :  -|-  is  the  same 
as  thai  of  8  : 4,  viz :  2.  Hence, 

488.  The  ratio  of  any  two  fractions  may  be  expressed  in 
whole  numbers,  by  reducing  them  to  a  common  denominator,  and 
then  using  the  numerators  for  the  terms  of  the  ratio.  (Art.  484.) 
Thus,  the  ratio  of  %  to  4-  is  the  same  as  -^  :  -j\,  or  6:2. 

33.  What  is  the  direct  ratio  of  4  to  12,  expressed  in  the  lowest 
terms?  Ans.  \. 

34.  What  is  the  inverse  ratio  of  4  to  12  ?     Ans.  -•L-j-TJ5=3. 

35.  What  is  the  direct  ratio  of  64  to  8  ?       Of  9  to  63  ? 

36.  What  is  the  direct  ratio  of  84  to  21  ?     Of  256  to  32  ? 

37.  What  is  the  inverse  ratio  of  4  to  16  ?     Of  28  to  7  ? 

38.  What  is  the  inverse  ratio  of  42  to  6  ?     Of  8  to  72  ? 

39.  Which  is  the  greater,  the  ratio  of  63  to  9,  or  that  of  72  to  8  ? 

40.  Which  is  the  greater,  the  ratio  of  86  to  240,  or  that  of  45 
to  72? 

41.  Which  is  the  greater,  the  ratio  of  120  to  85,  or  that  of 
240  to  170? 

42.  Which  is  the  greater,  the  ratio  of  624  to  416,  or  that  of 
936  to  560? 

43.  Is  the  ratio  of  5X6  to  24,  a  ratio  of  greater,  or  less  in 
equality? 

QUZST. — 486.  When  the  antecedent  is  greater  than  the  consequent,  whit  is  the  ratio 
called  ?  437.  If  the  antecedent  is  less  than  the  consequent,  what  is  the  ratio  called  1 
488  How  mav  the  ratio  of  two  fractions  be  expressed  in  whole  numbers  7 


318  RATIO.  \SECT.  XIV 

44.  Is  the  ratio  of  6  X  9  to  7  X  8,  a  ratio  of  greater,  or  less  in- 
equality ? 

45.  Is  the  ratio  of  2X^X16  to  4X32  a  ratio  of  greater,  or  less 
inequality  ? 

46.  What  is  the  ratio  compounded  of  the  ratios  of  5  to  3,  and 
12  to  4  ? 

47.  What  is  the  ratio  compounded  of  8  :  10,  and  20  :  16  ? 

48.  What  is  the  ratio  compounded  of  3  :  8,  and  10:5? 

49.  What  is  the  ratio  compounded  of  18  :  20,  and  30  :  40  ? 

50.  What  is  the  ratio  compounded  of  35  :  40,  and  60  :  75,  and 
21  tc  19? 

51.  What  is  the  ratio  compounded  of  60  :  40,  and  12  :  24,  and 
25:30? 

48 O«  In  a  series  of  ratios,  if  the  consequent  of  each  preced- 
ing couplet  is  the  antecedent  of  the  following  one,  the  ratio  of  the 
first  antecedent  to  the  last  consequent,   is  equal   to   that    com- 
pounded of  all  the  intervening  ratios. 
Thus,  in  the  series  of  ratios  3  :  4 

4:7 
7:  16 

the  ratio  of   3    to   16,  is  equal  to  that  which  is    compounded 
of  the  ratios   of  3  :  4,  of  4  :  7,  and   7  :  16;  for,  the   compound 

..     .    3X4X7 
ratio  is 

40O»  If  to  or  from  the  terms  of  any  couplet,  two  other  num 
lers  having  the  same  ratio  be  added  or  subtracted,  the  sums  or  re 
mainders  will  also  have  the  same  ratio.  (Thomson's  Legendre, 
B.  III.,  Prop.  1,  2.)  Thus,  the  ratio  of  12  :  3  is  the  same  as  that  of 
20  :  5.  And  the  ratio  of  the  sum  of  the  antecedents  12  +  20  to  the 
sum  of  the  consequents  3+5,  is  the  same  as  the  ratio  cf  either 
couplet.  That  is, 

12 +20:  3+5::  12:  3=20:  5,  or 

So  alsc  the  ratio  of  the  difference  of  the  antecedents,  to  the  dif- 
ference of  the  consequents,  is  the  same.  That  is, 

20—12     12     20 
20 — 12  :  5 — 3 : :  12  :  3=20  •  5,  or   5373-=^= ~^~-4- 


ARTS.  189-494.]  PROPORTION.  319 

49 1  •  If  in  several  couplets  the  ratios  are  equal,  the  sum  of 
all  tht  antecedents  has  the  same  ratio  to  the  sum  of  all  the  conse- 
quents, which  any  one  of  the  antecedents  has  to  its  consequent. 

(  12:4  =  3 

Thus,  the  ratio  of]  15:5  =  3 
(  18:6  =  3 
Therefore  the  ratio  of  (12  +  15  +  18) :  (4+5  +  6)=3. 

OBS.  1.  A  ratio  of  greater  inequality  is  diminished  by  adding  the  same  nurit- 
for  to  both  terms.  Thus,  the  ratio  of  8 : 2,  is  4 ;  and  the  ratio  of  8+4 :  2+ 
4  is  2. 

2.  A  ratio  of  less  inequality  is  increased  by  adding  the  same  number  to  both 
the  terms.  Thus,  the  ratio  of  2 : 8  is  i,  and  the  ratio  of  2+16 : 8+16  is  \. 

PROPORTION. 

492*  PROPORTION  is  an  equality  of  ratios.  Thus,  the  two 
ratios  6  :  3  and  4  :  2  form  a  proportion  ;  for  f =f ,  the  ratio  of  each 
being  2. 

OBS.  The  terms  of  the  two  couplets,  that  is,  the  numbers  of  which  the  pro- 
portion is  composed,  are  called  proportionals. 

493.  Proportion  may  be  expressed  in  two  ways. 

First,  by  the  sign  of  equality  (=)  placed  between  the  two 
ratios. 

Second,  by  four  points  (: :)  placed  between  the  two  ratios. 

Thus,  each  of  the  expressions,  12:6  =  4:2,  and  12  :  6  : :  4  :  2, 
is  a  proportion,  one  being  equivalent  to  the  other.  The  latter  ex- 
pression is  read,  "  the  ratio  of  12  to  6  equals  the  ratio  of  4  to  2,'* 
or  simply,  "  12  is  to  6  as  4  is  to  2." 

OBS.  The  sign  (: :)  is  said  to  be  derived  from  the  sign  of  equality,  thejow 
points  being  merely  the  extremities  of  the  lines. 

49  4»  The  number  of  terms  in  a  proportion  must  at  least  be 
four,  for  the  equality  is  between  the  ratios  of  two  couplets,  and 
each  couplet  must  have  an  antecedent  and  a  consequent.  (Art.  476.) 

There  may,  however,  be  a  proportion  formed  from  three  num- 
bers, for  one  of  the  numbers  may  be  repeated  so  as  to  form  two 

QUEST. — 492.  What  is  Proportion  ?  493.  How  many  ways  is  proportion  expressed  T 
What  is  the  first  ?  The  second  ?  494.  How  many  terms  must  there  be  in  a  proportion  1 
Why  1  Can  a  proportion  be  formed  of  three  numbers  7  How  ? 


320  PROPORTION.  [SECT.  XIV 

terms.  Thus,  the  numbers  8,  4,  and  2,  are  proportional  ;  for  the 
ratio  of  8  :  4=4  :  2.  It  will  be  seen  that  4  is  the  consequent  ia 
the  first  couplet,  and  the  antecedent  in  the  last.  It  is  therefore 
a  mean  proportional  between  8  and  2. 

OBS.  1.  In  this  case,  the  number  repeated  is  called  the  middle  term  or  mean 
proportional  between  the  other  two  numbers. 

The  last  term  is  called  a  third  proportional  to  the  other  two  numbers.  Thua 
2  is  a  third  proportional  to  8  and  4. 

2  Care  must  be  taken  not  to  confound  proportion  with  ratio.  (Arts.  474,  43*2.) 
In  a  simple  ratio  there  are  but  two  terms,  an  antecedent  and  a  consequent  , 
whereas  in  a  proportion  there  must  at  least  be  four  terms,  or  two  couplets. 

Again,  one  ratio  may  be  greater  or  less  than  another;  the  ratio  of  9  to  3  ia 
greater  than  the  ratio  of  8  to  4,  and  less  than  that  of  18  to  2.  QHC  proportion, 
on  the  other,  hand,  cannot  be  greater  or  less  than  another  ;  for  equality  doea 
not  admit  of  degrees. 

49  5  •  The  first  and  last  terms  of  a  proportion  are  called  the 
extremes  ;  the  other  two,  the  means. 

OBS.  Homologous  terms  are  either  the  two  antecedents,  or  the  two  conse- 
quents. Analogous  terms  are  the  antecedent  and  consequent  of  the  same 
couplet. 

496.  Direct  proportion  is  an  equality  between  two  direct 
ratios.     Thus,  12  :  4  :  :  9  :  3  is  a  direct  proportion. 

OBS.  In  a  direct  proportion,  the  first  term  has  the  same  ratio  to  the  second, 
as  the  third  has  to  the  fourth. 

497.  Inverse  or  reciprocal  proportion  is  an  equality  between 
a  direct  and  a  reciprocal  ratio.     Thus,  8  :  4  ::•£:•£-;  or  8  is  to  4, 
reciprocally,  as  3  is  to  6. 

OBS.  In  a  reciprocal  or  inverse  proportion,  the  first  term  has  the  same  ratio 
to  the  second,  as  the  fourth  has  to  the  third. 

498.  If  four  numbers  are  proportional,  the  product  of  the  ex- 
tremes is  equal  to  the  product  of  the  means.     Thus,  8  :  4  :  :  6  :  3  ' 
a  proportion;  for  f=f,  (Art.  492,)  and  8X3=4X6. 


.  —  Obs.  What  is  the  number  repeated  called  ?  What  is  the  last  term  called  ji 
such  a  case  ?  What  is  the  difference  between  proportion  and  ratio  ?  495.  Which  terms 
are  the  extremes  ?  Which  the  means  1  Obs-  What  are  homologous  terms?  Analrgous 
terms  ?  496.  What  is  direct  proportion  ?  Obs.  In  direct  proportion  what  ratio  has  the 
first  term  to  the  second  ?  497.  What  is  inverse  proportion1?  Oba.  What  ratio  has  the 
first  term  to  the  second  in  this  case  7  498.  If  four  numbers  are  proportional,  what  is  the 
puduct  of  the  extremes  equW  to? 


ARTS.  495-501.]  PROPORTION.  321 

Again,  12  :  6  :  :  i  :  •§•  is  a  proportion.  (Art.  496.) 
And      12xi=6xi 

OBS.  1.  The  truth  of  this  proposition  may  also  be  illustrated  in.  the  following 
manner  : 

The  numbers  2  :  3  :  :  6  :  9  are  obviously  proportional.  (Art.  492.) 

For,  f  =f.  (Art.  195.)    Now, 

Multiplying  each  ratio  by  27,  (the  product  of  the  denominators,) 


The  proportion  becomes?^??  =~--  (Art.2t.  Ax.  6. 

Dividing  both  the  numerator  and  the  denominator  of  the  first  couplet  by  3, 
(Art.  191,)  or  canceling  the  denominator  3  and  the  same  factor  in  27,  (Art.  221,) 
also  canceling  the  9,  and  the  same  factor  in  27,  we  have  2x9=6x3.  But  2 
and  9  are  the  extremes  of  the  given  proportion,  and  3  and  6  are  the  means  ;, 
hence,  the  product  of  the  extremes  is  equal  to  the  product  of  the  means. 

2.  Conversely,  if  the  product  of  the  extremes  is  equal  to  the  product  of  the 
means,  the  four  numbers  are  proportional  ;  and  if  the  products  are  not  equal, 
the  numbers  are  not  proportional. 

499,  Proportion,  ii*  arithmetic,  is  usually  divided  into  Simple 
and  Compound. 

SIMPLE  PROPORTION. 

500,  SIMPLE  PROPORTION  is  an  equality  between  two  simple 
ratios.     It  may  be  either  direct  or  inverse.  (Arts.  479,  496,  497.) 

The  most  important  application  of  simple  proportion  is  the 
solution  of  that  class  of  examples  in  which  three  terms  are  given  to 
find  a  fourth. 

501,  We  have  seen  that,  if  four  numbers  are  in  proportion, 
the  product  of  the  extremes  is  equal  to  the  product  of  the  means. 
(Art.  498.)     Hence, 

If  the  product  of  the  means  is  divided  by  one  of  the  extremes, 
the  quotient  will  be  the  other  extreme  ;  and  if  the  product  of  the 
extremes  is  divided  by  one  of  the  means,  the  quotient  will  be  the 

QUEST.  —  Obs.  If  the  product  of  the  extremes  is  equal  to  the  product  of  the  means,  what 
le  true  of  the  four  numbers  1  If  the  products  are  not  equal,  what  Is  true  of  them  ?  499 
How  is  proportion  usually  divided  1  500.  What  is  simple  proportion  7  What  is  the  most 
Important  application  of  it?  501.  If  the  product  of  the  means  is  divided  by  one  of  th« 
extremes,  what  will  the  quotient  be  ?  If  the  product  of  the  extremes  is  divided  by  one  of 
the  means,  what  will  the  quotient  be  ? 


322  PROPORTION  [SECT.  XIV 

other  mean.     For,  if  the  product  of  two  factors  is  divided  by  one 
of  them,  the  quotient  will  be  the  other  factor.  (Art.  156.) 

Take  the  proportion  8  :  4  :  :  6  :  3. 

Now  the  product       8X3  —  4  =  6,  one  of  the  means  ; 

So  the  product          8x3  —  6=4,  the  other  mean. 

Again,  the  product   4x6—8=3,  one  of  the  extremes; 

And  the  product       4X6—3  =  8,  the  other  extreme. 

5  O2«  If,  tJierefore,  any  three  terms  of  a  proportion  are  given, 
the  fourth  may  be  found  by  dividing  the  product  of  two  of  tJiem 
by  the  other  term. 

OBS.  Simple  Proportion  is  often  called  the  Rule  of  TTirce,  from  the  circum- 
stance that  three  terms  are  given  to  find  a  fourth.  In  the  older  arithmetics,  it 
is  also  called  the  Golden  Ride.  But  the  fact  that  these  names  convey  no  idea 
of  the  nature  or  object  of  the  rule,  seems  to  be  a  strong  objection  to  their  use, 
not  to  say  a  sufficient  reason  for  discarding  them. 

Ex.  1.  If  the  product  of  the  means  is.  84,  and  one  of  the  ex- 
tremes is  7,  what  is  the  other  extreme,  or  term  of  the  proportion  ? 

2.  If  the  product  of  the  means  is  54,  and  one  of  the  extremes 
is  18,  what  is  the  other  extreme  ? 

3.  If  the  product  of  the  means  is  720,  and  one  of  the  extremes 
is  45,  what  is  the  other  extreme  ? 

4.  If  the  product  of  the  means  is  639,  and  one  of  the  extremes 
is  213,  what  is  the  other  extreme  ? 

5.  If  the  first  three  terms  of  a  proportion  are  8,  12,  and  16, 
is  the  fourth  term  ? 


Solution.  —  12X16  =  192,  and  192-r8  =  24,  the  fourth  term,  or 
number  required  ;  that  is,  8  :  1  2  :  :  16  :  24. 

6.  It  is  required  to  find  the  fourth  term  of  the  proportion,  the 
first  three  terms  of  which  are  36,  30,  and  24. 

7.  Required  the  fourth  term  of  the  proportion,  the  first  three 
terms  of  which  are  15,  27,  and  31. 

8.  Required  the  fourth  term  of  the  proportion  whose  first  three 
terms  are  45,  60,  and  90. 

QUEST.—  Obs.  What  is  simple  proportion  often  called  ?    Do  these  terms  convey  an  /deft 
of  the  nature  or  object  of  the  rule  1 


ARTS.  502, 503.J  PROPORTION.  323 

9.  If  8  yds.  of  broadcloth  cost  $96,  how  much  will  20  yds.  cost 
at  the  same  rate  ? 

Solution. — It  is  plain  that  8  yds.  has  the  same  ratio  to  20  yds. 
as  the  cost  of  8  yds.,  viz :  $96,  has  to  the  cost  of  20  yds.  That  is, 

8  yds. :  20  yds. : :  $96  :  to  the  cost  of  20  yds. 
Now  $96X20— $1920;  and  $1920-r8=$240.  Ans. 

10.  If  35  men  will  consume  a  certain  quantity  of  flour  in  20 
lays,  how  long  will  it  take  50  men  to  consume  it  ? 

Note.—  Since  the  answer  is  days,  we  put  the  given  days  for  the  third  term. 
Then,  as  the  flour  will  not  last  50  men  so  long  as  it  will  35  men,  we  put  the 
smaller  number  of  men  for  the  second  term,  and  the  larger  for  the  first. 

Operation. 

Men.     Men.        Days. 

50  :  35  :  :  20  :  to  the  number  of  days  required. 

20  Multiply  the  second  and  third  terms  to- 

50  )  700  gether,  and  divide  the  product  by  the  first 

]  4  days.  Ans.     term,  as  in  the  last  example. 
PROOF.— 50X14-35X20.  (Art.  498.) 

5O3.  From  the  preceding  illustrations  and  principles,  we  de- 
duce the  following  general 

RULE  FOR   SIMPLE  PROPORTION. 

I.  Place  that  number  for  the  third  term,  which  is  of  the  same 
kind  as  the  answer  or  number  required. 

11.  Then,  if  by  the  nature  of  the  question  the  answer  must  be 
greater  than  the  third  term,  place  the  greater  of  the  other  two  num- 
bers for  the  second  term  ;  but  if  it  is  to  be  less,  place  the  less  of  the 
other  two  numbers  for  the  second  term,  and  the  other  for  the  first. 

III.  Finally,  multiply  the  second  and  third  terms  together,  divide 
the  product  by  the  first,  and  the  quotient  will  be  the  answer  in  the 
same  denomination  as  the  third  term. 

PROOF. — Multiply  the  first  term  and  the  answer  together,  and 
if  the  product  is  equal  to  the  product  of  the  second  and  third  terms, 
the  work  is  right.  (Art.  500.) 

QUEST. — 503.  In  arranging  the  terms  in  simple  proportion,  which  number  is  pnt  for  the 
third  term  ?  How  arrange  the  other  two  numbers  ?  Having  stated  the  question  how  is  the 
ans  wer  found  1  Of  what  U  ^nomination  is  the  answer  ?  How  is  simple  proportion  proved  I 


324  SIMPLE  [SECT.  XIV. 

Demonstration,  -If  four  numbers  are  proportional,  we  have  seen  that  the 
product  of  the  means  is  equal  to  the  product  of  the  extremes;  (Art.  498;)  there- 
fore the  poduct  of  the  second  and  third  terms  must  be  equal  to  that  of  the  first 
and  fourth.  But  if  the  product  of  two  factors  is  divided  by  one  of  them,  the 
quotient  will  be  the  other;  (Art.  15G ;)  consequently,  when  the  first  three 
lerms  of  a  proportion  are  given,  the  product  of  the  second  and  third  terms  di- 
vided by  the  first,  must  give  the  fourth  term  or  answer. 

The  object  of  placing  that  number,  which  is  of  the  same  kind  as  the  answer, 
for  the  third  term,  instead  of  the  second,  as  is  sometimes  done,  is  twofold  :  1st, 
it  avoids  the  necessity  of  the  Rale  of  Three  Inverse ;  2d,  the  third  term,  in 
many  cases,  has  no  ratio  to  the  first ;  consequently  it  is  inconsistent  with  tfaa 
principles  of  proportion  to  put  it  for  the  second  term.  Thus,  in  the  ninth  ex- 
ample, if  we  put  $96  for  the  second  term,  it  would  read,  8  yds.  :  $9G : :  20 
yds. :  $240,  the  answer.  But  a  yard  can  have  no  ratio  to  a  dollar ;  for  one 
Cannot  be  said  to  be  greater  nor  less  than  the  other.  (Art.  476.  Obs.  2.) 

OBS.  1.  If  the  first  and  second  terms  are  compound  numbers,  reduce  them 
to  the  lowest  denomination  mentioned  in  either,  before  the  multiplication  or 
division  is  performed. 

When  the  third  term  contains  different  denominations,  it  must  also  be  re- 
duced to  the  lowest  denomination  mentioned  in  it. 

2.  The  process  of  arranging  the  terms  of  a  question  for  solution,  or  put- 
ting it  into  the  form  of  a  proportion,  is  called  stating  the  question. 

3.  Questions   in   Simple   Proportion,   we  have  seen,   may   be  solved    by 
Analysis.     After  solving  the  following  examples  by  proportion,  it  will  be  an  ex- 
•ellent  exercise  for  the  student  to  solve  them  by  analysis.    (Art.  4G2.  Obs.  2.) 

11.  If  16  barrels  of  flour  cost  $112,  what  will  129  barrels  cost? 

12.  If  40  acres  of  land  cost  $540,  what  will  97  acres  cost? 

13.  If  641  sheep  cost  $1923,  what  will  75  sheep  cost? 

14.  At  the  rate  of  155  miles  in  12  days,  how  far  can  a  man 
travel  in  60  days  ? 

15.  How  much  hay,  at  $17.50  per  ton,  can  you  buy  for  $350  ? 

16.  If  $45  buy  63  Ibs.  of  tea,  how  much  will  $1540  buy? 

17.  If  90  Ibs.  of  pepper  are  worth  72  Ibs.  of  ginger,  how  many 
Ibs.  of  ginger  are  64  Ibs.  of  pepper  worth  ? 

18.  A  bankrupt  compromised  with  his  creditors,  at  64  cts.  on  a 
dollar ;  how  much  will  be  received  on  a  debt  of  $2563.50  ? 

19.  An  emigrant  has  a  draft  for  £1460  sterling:  how  much  is 
it  worth,  allowing  $4.84  to  a  pound  ? 

2»7«ST.— Obs.  [f  the  first  and  second  terms  contain  different  denominate  ns,  how  pro- 
ceed 7  When  the  it  ird  term  contains  different  denominations,  what  is  to  be  >nne  1  What 
b  meant  oy  stating  a  question  1 


ART.  504. J  PROPORTION.  325 

SIMPLE   PROPORTION  BY  CANCELATION. 
20.  If  72  tons  of  coal  cost  $648,  how  much  will  9  tons  cost? 
Operation.  Having  stated  the  question  as  be« 

T-?? '  -T<TS-  •  D°llS      A   <•        *°re>  we  Perce^ve  tlie  factor  9  is  com- 
mon  to  the   first   two   terms,   and 

therefore  may   be    canceled.    (Art. 

Now  $648-|-8=$81.  Ans.        _K1v  J  V 

iDi.j 

Or  thus,  -^ — =the  answer.  (Art.  503.) 

9Vfi4-8      d)  y  648 
But     — - — =~TA — =$81,  the  same  as  before.     Hence, 

72  *£>8 

5O4.  When  the  first  term  has  factors  common  to  either  of 
the  other  two  terms. 

Cancel  the  factors  which  are  common,  then  proceed  according  to 
the  rule  above.  (Arts.  151,  221.) 

PROOF. — Place  the  answer  in  the  denominator,  or  on  the  left  of 
the  perpendicular  line,  as  the  case  may  be,  and  if  the  factors  of  the 
divisor  exactly  cancel  those  of  the  dividend,  the  work  is  right. 

OBS.  1.  The  question  should  be  stated,  before  attempting  to  cancel  the  com- 
mon factors.  When  the  terms  are  of  different  denominations,  the  reduction 
of  them  may  sometimes  be  shortened  by  Cancelation. 

2.  Instead  of  points,  it  may  sometimes  be  more  convenient  fo  place  a  per- 
pendicular line  between  the  first  and  second  terms,  as  in  division  of  fractions. 
(Art.  231.)     In  this  case  the  third  term  should  be  placed  under  the  second, 
with  the  sign  of  proportion  (  : :  )  before  it  to  denote  its  origin,  and  its  relation 
n  the  fourth  term  or  the  answer. 

3.  It  will  be  perceived  that  cancelation  is  applicable  in  Simple  Proportion  to 
all  those  examples;  whose  first  term  has  one  or  more  factors  common  to  either 
of  the  other  terms. 

21    If  24  yds.  of  cloth  cost  $63,  what  will  32  yds.  cost  ? 
Operation. 


yds. 


20  yds.,  40  When  arranged  in  this  way,  th 

::$$$,  21  question  is  read,  24  yds.  is  to  320 


Ans.  |$21  X  40=$840.     yds.,  as  $63  is  to  the  answer  required. 
22.  If  20  bu.  of  oats  cost  £l,  how  much  will  2  quarts  cost  ? 


326  SIMPLE  [SECT.  XIV 

23.  If  12  bbls.  of  flour  cost  $88,  what  will  108  barrels  cost9 

24.  If  30  cows  cost  $480,  what  will  173  cows  cost  ? 

25.  If  a  man  can  travel  240  miles  in  16  days,  how  far  can  he 
travel  in  29  days? 

26.  If  48  men  can  build  a  ship  in  84  days,  how  long  would  it 
take  1 6  men  to  build  it  ? 

27.  If  ^  of  a  ton  of  hay  costs  £-£,  what  will  |  of  a  ton  cost  ? 

ton.  ton.       £ 

Solution. — J- :  -I  :  :  f  :  Ans.  Now,f  XlXi=£2  Ans.   Hence, 

5O5.  If  the  terms  in  a  proportion  are  fractional,  the  question 
is  stated,  and  the  answer  obtained  in  the  same  manner  as  if  they 
were  whole  numbers. 

OBS.  When  the  first  and  second  terms  are  fractions,  we  may  reduce  them  to 
a  common  denominator,  and  then  employ  the  numerators  only ;  for  the  ratio 
of  two  fractions  which  have  a  common  denominator,  is  the  same  as  the  ratio 
of  their  numerators.  (Art.  487.  Obs.  2.) 

28.  If  f of  a  cord  of  wood  cost  $1.35,  what  will  f  of  a  cord 
cost? 

29.  If  •£  of  a  yard  of  berege  cost  6  shillings,  what  will  $  of  a 
yard  cost  ? 

30.  If  f  of  a  yard  of  sarcenet  cost  f  of  a  dollar,  what  will  3| 
yds.  cost? 

31.  If  |  of  a  pound  of  chocolate  cost  |  of  a  dollar,  what  will 
25f  pounds  cost  ? 

32.  What  will  165  melons  cost,  at  $•  of  a  dollar  for  5  melons? 

33.  A  man  had  420  acres  of  land  which  he  wished  to  divide 
among  his  three  sons  A,  B,  and  C,  in  proportion  to  the  numbers 
7,  5,  and  3  :  how  much  land  would  each  receive  ? 

Solution. — Since  the  several  parts  are  to  be  proportional  to  the 
numbers  7,  5,  and  3,  the  sum  of  which  is  15,  it  is  evident  that  the 
Bum  of  all  the  given  numbers  is  to  any  one  of  them,  as  the  whole 
quantity  to  be  divided  to  the  part  corresponding  to  the  number 
used  as  the  second  term. 

That  is  15  :  7  : :  420 A.  to  A's  share,  which  is  196  acres; 

Also      15:5::  420A.  to  B's      "          "      "  140  acres ; 

And       15  :  3  : :  420A.  to  C's      "          "      "    84  acres. 

PROOF.—  196+140+84=420A.  the  given  number.   (Ax.  11.) 


ARTS   505, 506.]  PROPORTION.  327 

5O6.  Hence,  to  divide  a  given  number  or  quantity  into  parts 
which  shall  be  proportional  to  any  given  numbers. 

Place  the  whole  number  or  quantity  to  be  divided  for  the  third 
term,  the  sum  of  the  given  numbers  for  the  first  term,  and  each  of 
the  given  numbers  respectively  for  the  second;  then  multiply  ard 
divide  as  before.  (Art.  503.) 

31  A  farmer  wishes  t:  mix  100  bushels  of  provender  of  oats 
and  corn  in  the  ratio  of  3  to  7 :  how  many  bushels  of  each  must 
he  put  in  ? 

35.  Bell  metal  is  composed  of  3  parts  of  copper,  and  1  of  tin : 
how  much  of  each  ingredient  will  be  used  in  making  a  bell  which 
weighs  2567  pounds  ? 

36.  Gunpowder  is  composed  of  76  parts  of  nitre,  14  of  char- 
coal, and  10  of  sulphur :  how  much  of  each  of  these  ingredients 
will  it  take  to  make  a  ton  of  powder  ? 

37.  If  40.12  Ibs.  of  sugar  are  worth  $5.13,  how  much  can  be 
bought  for  $125.375  ? 

.38.  The  Vice-President's  salary  is  $5000  a  year:  if  his  daily 
expenses  are  $10,  how  much  can  he  lay  up  ? 

39.  If  f  Ib.  of  snuff  cost  £fc,  what  will  150  Ibs.  cost? 

40.  If  -f  of  f  of  f  of  a  sloop  cost  $1500,  what  will  the  whole  cost? 

41.  If  -£  of  -f-  of  an  acre  of  land  on  Broadway  is  worth  $8200, 
how  much  is  -£  of  -f*of  an  acre  worth  ? 

42.  A  man  bought  •£  of  a  vessel  and  sold  f  of  what  he  bought 
for  $8240,  which  was  just  the  cost  of  it :  what  was  the  whole 
vessel  worth  ? 

43.  How  many  times  will  the  fore  wheel  of  a  carriage  which  is 
7  ft.  6  in.  in  circumference  turn  round  in  going  100  miles  ? 

44.  How  many  times  will  the  hind  wheel   of  a  carriage  9  ft. 
2  in.  in  circumference,  turn  round  in  going  the  same  distance  ? 

45.  There  are  two  numbers  which  are  to  each  other  as  12  to 
84,  the  smaller  of  which  is  75  :  what  is  the  larger? 

46.  What  two  numbers  are  those  which  are  to  each  other  as 
6  to  6,  the  greater  of  which  is  240  ? 

47.  If  two  numbers  are  as  8  to  12,  and  the  less  is  320,  what 
is  the  greater  ? 


378  PROPORTION.  [SECT.  XIV. 

48.  There  are  two  flocks  of  sheep  which  are  to  each  other  as  15 
to  20,  and  the  greater  contains  500  :  how  many  does  the  less  con- 
tain ? 

49.  An  express  traveling  60  miles,  a  day  had  been  dispatched 
5  days,  when  a  second  was  sent  after  him  traveling  75  miles  a 
day  :  how  long  will  it  take  the  latter  to  overtake  ,ne  former  ? 

50.  A  foi  has  150  rods  the  start  of  a  hound,  but  the  hound 
tuns  6  rods  while  the  fox  runs  5  rods  :  how  far  must  the  hound 

un  before  he  catches  the  fox  ? 

51.  A  stack  of  hay  will  keep  a  cow  20  weeks,  and  a  horse  15 
weeks  :  how  long  will  it  keep  them  both  ? 

52.  A  traveler   divided  80s.  among  4  beggars  in  such  a  man- 
ner, that  as  often  as  the  first  received  10s.,  the  second  received 
f>s.,  the  third  8s.,  and  the   fourth  7s. :  what  did  each  receive  ? 

53.  Pure  water  is  composed  of  oxygen  and  hydrogen  in  the  ratic 
of  8  to  1  by  weight :  what  is  the  weight  of  each  in  a  cubic  foot 
of  tvater,  or  1000  ounces  avoirdupois? 

COMPOUND   PROPORTION. 

5O7  COMPOUND  PROPORTION  is  an  equality  between  a  com- 
pound  ratio  and  a  simple  one.  (Arts.  479,  480.) 

us'      *     >  : :  12  :  3,  is  a  compound  proportion. 
Into     4:2$ 

That  is,  6X4  :  3X2  : :  12  :  3  ;  for,  6X4X3  =  3X2X12. 

OBS.  Compound  proportion  is  chiefly  applied  to  the  solution  of  exe.mj  let 
which  would  require  two  or  more  statements  in  simple  proportion.  It  is  some- 
times called  Double  Ruk  of  Three. 

Ex.  1.  If  8  men  can  reap  32  acres  in  6  days,  how  many  acres 
can  12  men  reap  in  15  days? 

Suggestion. — When  stated  in  the  form  of  a  compound  propor- 
tion, the  question  will  stand  thus : 

8m.  :  12m.  i 

.  ,       ,  K      S  : :  32  A.  : to  the  answer. 

6d.  :  15a.   > 

That  is,  the   product  of  the  antecedents  8X6,  has  the  same 

QUEST.— 507.  What  is  compound  proportion  1  Ob*.  To  what  Is  it  cAiefl-  rpplled  f 
What  is  it  sometimes  called  1 


ARTS.  507,  508.]  PROPORTION.  329 

ratio  tc  the  product  of  the  consequents  12x15,  as  32  has  to  the 
answer ;   o,  simply,  8  into  6:12  into  15  : :  32  :  to  the  answer. 

Operation.  The  product  of   the  numbers 

32X12X15  =  5760,  standing  in  the  2d  and  3d  places 

And             8X   6  =  48.  divided  by  the  product  of  those 

Now      5760^-48=120.  standing  in  the    first  place,  will 

Ans.  12,0  acres.       give  the  answer. 

• 

Not..  —The  learner  will  observe  that  it  is  not  the  ratio  of  8  to  12  alone, 
nor  that  of  6  to  15,  which  is  equal  to  the  ratio  of  32  to  the  answer,  as  it  it 
sometimes  stated  ;  but  it  is  the  ratio  compounded  of  8  to  12,  and  6  to  15,  which 
is  equal  to  the  ratio  of  32  to  the  answer.  Thus,  8X6  :  12X15 : :  32  :  120,  tha 
answer.  A  compound  proportion  when  stated  as  above,  is  read, "  the  ratio  of 
8X6  is  to  12X15  as  32  is  to  the  answer." 

2  If  6  men  can  earn  £42  in  60  days  working  8  hours  per  day, 
how  much  can  10  men  earn  in  84  days  working  12  hours  a  day? 

Operation. 

State  the  question,  then 
6m.    :  10m.      \  5     ,.  . , 

60d     •  84d        (  : :  £42  :  to  Ans.      mult'P'y   and    dlvlde>   as 

V  before. 

8hrs.  :  12hrs.    ) 

10X84X12X42=423360;  and  6X60X8=2880. 
Now       423360H-2880=147.     Ans.  £147. 

5O8.  From  the  foregoing  illustrations  we  derive  the  follow- 
ing general 

RULE   FOR  COMPOUND  PROPORTION. 

I.  Place  that  number  which  is  of  the  same  kind  as  the  answer 
required  for  the  third  term. 

II.  Then  take  the  other  numbers  in  pairs,  or  two  of  a  kind,  and 
arrange  them  as  in  simple  proportion.  (Art.  503.) 

III.  Finally,  multiply  together  all  the  second  and  third  termst 
divide  the  result  by  the  product  of  the  first  terms,  and  the   quo- 
tient will  be  the  fourth  term  or  answer  required. 

QUEST. — 508.  In  stating  a  question  in  compound  proportion,  which  number  do  you  put  fot 
the  third  term  1  How  arrange  the  other  numbers  ?  Having  stated  the  question,  how  i| 
the  answer  found"? 


330  COMPOUND  [SECT.  XIV. 

PROOF. — Multiply  the  answer  into  all  of  the  first  terms  or  ante- 
cedents of  the  first  couplets,  and  if  the  product  is  equal  to  the  con- 
tinued product  of  all  the  second  and  third  terms,  the  work  is  right. 
(Art.  498.) 

OBS.  1.  Among  the  given  numbers  there  is  but  one  which  is  of  the  same 
kind  as  the  answer.  This  is  sometimes  called  the  odd  terrn^  and  is  always  to 
be  placed  for  the  third  term. 

2.  If  the  antecedent  and  conseqiienl  of  any  couplet  are  compound  numbers, 
they  must  be  reduced  to  the  lowest  denomination  mentioned  in  either,  before 
Ihe  multiplication  is  performed.     When  the  third  term  contains  different  de- 
nominations, it  must  also  be  reduced  to  the  lowest  mentioned  in  it. 

3.  Questions  in  Compound  Proportion  may  be  solved  by  Analysis ;  also  by 
Simple  Proportion,  by  making  two  or  more  separate  statements. 

3.  If  12  horses  can  plough  11   acres  in  5  days,  how  many 
horses  can  plough  33  acres  in  18  days? 

4.  If  a  man  walking  12  hours  a  day,  can  travel  250  miles  in  10 
days,  how  long  will  it  take  him  to  travel  400  miles,  if  he  walks  but 
10  hours  a  day  ? 

5.  If  40  gallons  of  water  will  last  20  persons  5  days,  how 
many  gallons  will  9  persons  drink  in  a  year  ? 

6.  If  16  laborers  can  earn  £15,  12s.  in  18  days,  how  many 
laborers  will  it  take  to  earn  £35,  2s.  in  24  days  ? 

COMPOUND   PROPORTION  BY  CANCELATION. 

7.  If  a  person  can  make  60  rods  of  wall  in  45  days,  working  12 
hours  a  day,  how  many  rods  can  he  make  in  72  days,  working  8 
hours  a  day  ? 

Statement. 

45d.       :  72d.      )     Rods- 
12hrs.    :    8hrs.    [:'•  60  :  to  the  answer.     That  is, 

0,2         4 

=  64  rods.  Ans.     Hence, 


45X12 


QUEST. — How  are  questions  in  compound  proportion  proved  ?  Obs.  Among  the  given 
numbers,  how  many  are  of  the  same  kind  as  the  answer  1  Can  questions  in  compound 
proportion  be  solved  in  any  other  way  ? 


ART.  509.]  PROPORTION.  331 

5OO.  When  the  first  terms  have  factors  common  to  the  sec- 
ond or  third  terms. 

Cancel  the  factors  which  are  common,  then  div  de  the  product  of 
those  remaining  in  the  second  and  third  terms  by  the  product  of 
those  remaining  in  the  first,  and  the  quotient  will  be  the  answer. 

PROOF. — Place  the  answer  in  the  denominator,  or  on  the  left  of 
the  perpendicular  line,  and  if  the  factors  of  the  divisor  and  dividend 
exactly  cancel  each  other,  the  work  is  right. 

OBS.  1.  Instead  of  placing  points  between  the  antecedents  and  consequents 
of  the  left  hand  couplets  of  the  proportion,  it  is  sometimes  more  convenient  to 
put  a  perpendicular  line  between  them,  as  in  division  of  fractions.  (Art.  23<J.J 
T.bis  will  bring  all  the  terms  whose  product  is  to  be  divided  on  the  right  of  the 
line,  and  those  whose  product  is  to  form  the  divisor,  on  the  left.  In  this  case 
the  third  term  should  be  placed  below  the  second  terms,  with  the  sign  of  pro- 
portion (: :)  before  it,  to  show  its  origin,  and  its  relation  to  the  answer. 

2.  It  will  be  observed  that  Cancelation  can  be  applied  in  Compound  Pro- 
portion to  all  those  examples  whose^rs^  terms  have  factors  "common  to  the 
second  terms,  or  to  the  third  term. 

8.  If  24  men  can  saw  90  cords  of  wood  in  6  days,  when  the 
days  are  9  hours  long,  how  many  cords  can    8   men  saw  in  36 
days,  when  they  are  1 2  hours  long  ? 
Operation. 

fa. 

,.  >d.,  2 
0hrs.  12hrs. 

:  00c.,  10 


Ans.     2X12X10  =  240  cords. 

9.  If  6  men  can  make  120  pair  of  boots   in  20  days,  working 
8  hours  a  day,  how  long  will  it  take  12  men  to  make  360  pair, 
working  10  hours  a  day? 

10.  If  12  men  can  build  a  wall  30  ft.   long,  6  ft.  high,  and    ' 
ft.  thick,  in  18  days,  how  long  will  it  take  36  men  to  build  on. 
360  ft.  long,  8  ft.  high,  and  6  ft.  thick. 

11.  If  a  horse  can  travel  120  miles  in  4  days  when  the  day 
are  8  hours  long,  how  far  can  he  travel  in  30  days  when  the  days 
are  10  hours  loner  ? 


. — 509.  When  the  first  terms  have  factors  conuncn  to  the  second  or  third  terma, 
foow  proceed  ? 

15 


>w  pr 
T.1J 


332  CONJOINED  [SECT.  XIV, 

12.  If  $250  gam  $30  in  2  years,  what  will  be  the  interest  of 
$750 'for  5  years? 

13.  What  will  be  the  interest  of  $500  for  4  years,  if  $600  will 
gain  $42  in  1  year  ? 

14.  If  $360  gain  $14.40  in  8  months,  what  will  $4800  gain  in 
32  months? 

15.  If  a  family  of  8   persons  spend  $200  in  9  months.,  how 
much  will  18  persons  spend  in  12  months? 

16.  If  15  men,  working  12  hours  a  day,  can  hoe  60  acres  in  20 
days,  how  long  will  it  take  30  boys,  working  10  hours  a  day,  to 
hoe  96  acres,  6  men  being  equal  to  10  boys? 

CONJOINED    PROPORTION. 

5  1 0«  When  each  antecedent  of  a  compound  ratio  is  equal  in 
value  to  its  consequent,  the  proportion  is  called  Conjoined  Propor- 
tion. 

OBS.  Conjoined  Proportion  is  often  called  the  chain  rule.  It  is  chiefly  used 
in  comparing  the  coins,  weights  and  measures  of  two  countries,  through  the 
medium  of  those  of  other  countries,  and  in  the  higher  operations  of  ex- 
change. The  odd  term  is  sometimes  called  the  demand. 

17.  If  20  Ibs.  United  States  make  12  Ibs.  in  Spain  ;  and  15  Ibs. 
Spain  20 Ibs.  in  Denmark;  and  40 Ibs.  Denmark  60  Ibs.  in  Russia: 
how  many  pounds  in  Russia  are  equal  to  100  Ibs.  U.  S.  ? 

Operation.  Arrange  the  given  terms  in 

20  Ibs.  U.  S.  =  12  Ibs.  Spain      pairs,  making  the  first  term  the 

15  Ibs.  Spain— 20  Ibs.  Den.       antecedent,  and  its  equal  the 

40  Ibs.  Den.  =60  Ibs.  Rus.        consequent;    then   since  it  is 

How  many  Ibs.  R.  =  100  Ibs.  U.  S.     required  to  find  how  many  of 

the   last  kind  are  equal  to  a 

given  number  (100  Ibs.)  of  the  first,  place  the  odd  term  or  de- 
mand under  the  consequents. 

Then,  20X15X40  :  12X20X60  : :  100  :  Ans. 
That  is       #vt|     12  Cancel  the  factors  common 

20  to  both  sides,  and  the  prodv  ft 


10, 


Ans. 


00,  JL  of  those  remaining  on  the  right. 

::100, 10  divided  by  the  product  of  those 


1 2  X 1 0 = 1 2 0  Ibs.     on  the  left,  is  t  he  answer. 


AllTrf.   510,  511.]  PROPORTION.  333 

511*  From  these  illustrations  we  derive  the  following 
RULE   FOR   CONJOINED   PROPORTION. 

I.  Taking  the  terms  in  pairs,  place  the  first  term  on  the  left  of 
the  sign  of  equality  or  a  perpendicular  line  for  the  antecedent,  and 
its  equal  on  the  right  for  the  consequent,  and  so  on.      Then,  if  the 
answer  is  to  be  of  the  same  kind  as  the  first  term,  place  tlie  odd 
term  under  the  antecedents  j  but  if  not,  place  it  under  the  conse- 
quents. 

II.  Cancel  the  factors  common  to  both  sides,  and  if  the  odd  term 
fells  under  the  consequents,   divide  the  product  of  the  factors  re- 
maining on  the  right  by  the  product  of  those  on  the  left,  and  the  quo- 
tient will  be  the  answer  ;  but  if  the  odd  term  falls  under  tlie  ante- 
cedents, divide  the  product  of  tlie  factors  remaining  on  the  left  by 
the  product  of  those  on  the  right,   and    the    quotient  will  be  the 
answer. 

PROOF. — Reverse  the  operation,  taking  the  consequents  for  the 
antecedents,  and  the  answer  for  the  odd  term,  and  if  the  result  thus 
obtained  is  the  same  as  the  odd  term  in  the  given  question,  the  work 
is  right. 

OBS.  In  arranging  the  terms,  it  should  be  observed  that  the  Jifst  antecedent 
and  the  Last  consequent  will  always  be  of  the  same  kind. 

18.  If  100  Ibs.  United  States,  make  95  Ibs.  Italian;  and  19  Ibw, 
Italian,  25  Ibs.  in  Persia ;  how  many  pounds  in  the   U.  S.  are 
equal  to  50  Ibs.  in  Persia  ?  Ans.  40  Ibs. 

19.  If  10  yds.  at  New  York  make  9  yds.  at  Athens;  and  90 
yds.  at   Athens,  112  yds.   at  Canton;  how  many  yds.  at  Canton 
are  equal  to  50  yds.  at  New  York  ? 

20.  If  50  yds.  of  cloth  in  Boston  are  worth  45  bbls.  of  flour  in 
Philadelphia;  and  90  bbls.  of  flour  in  Philadelphia  127  bales  of 
cotton  in  New  Orleans ;  how  many  bales   of  cotton  at  New  Or- 
leans are  worth  100  yds.  of  cloth  in  Boston? 

21.  If  $18  U.  S.  are  worth  8  ducats  at  Frankfort;  12  ducat? 
at  Frankfort  9  pistoles  at  Geneva ;  and  50  pistoles  at  Geneva,  t? 
rupees  at  Bombay :  how  many  rupees  at  Bombay  are  equal  to 
$100  United  States? 


334  DUODECIMALS.  [SECT.   XV. 

SECTION     XV 
DUODECIMALS. 

ART.  512*  DUODECIMALS  are  a  species  of  compound  numbers, 
the  denominations  'of  which  increase  and  decrease  uniformly  in  a 
twelvefold  ratio.  The  denominations  are  feet,  inches  or  primes, 
seconds,  thirds,  fourths,  fiftlis,  tfec. 

Note.  —  The  term  duodecimal  is  derived  from  the  Latin  numeral  du^)decimi 
which  signifies  twelve. 

TABLE. 

12  fourths       ("")     make  1  third,  marked  "' 

12  thirds  "  1  second,  "        " 

12  seconds  "  1  inch  or  prime,         "        in.  or  ' 

1  2  inches  or  primes      "  1  foot,  "       //. 

Hence  1'  =-,V  of  1  foot. 

1"  =^  of  1  in.,  or  -^  of  -f\  of  1  H.=T±T  of  1  ft. 
l'"=-iV  of  1",  or  A  of  -i  of  tV  of  1  ft.  =7^  of  1  ft. 


OBS.  The  accents  used  to  distinguish  the  different  denominations  below  feet, 
are  called  Indices. 

513*  Duodecimals  may  be  added  and  subtracted  in  the  same 
manner  as  the  other  compound  numbers.  (Arts.,  300,  302.) 

MULTIPLICATION   OF   DUODECIMALS. 

514.  Duodecimals  are  principally  applied  to  the  measurement 
of  surfaces  and  solids.  (Arts.  285,  286.) 

Ex.  1.  How  many  square  feet  are  there  in  aboard  12  ft.  7  in. 
long,  and  4  ft.  3  in.  wide  ? 

QUEST.—  512.  What  are  duodecimals  ?  What  are  the  denominations?  Note.  What  is 
the  meaning  of  the  term  duodecimal  ?  Repeat  the  Table.  Obs.  What  are  the  accent* 
called,  which  are  used  to  distinguish  the  different  denominations  1  513.  How  ar<  duodeci- 
mals added  and  subtracted  ?  514.  To  what  are  duodecimals  chiefly  appliod  ? 


ARTS.  512-515.]  DUODECIMALS  335 


Operation.  ^  e  ^rs*  multiply  each  denomination  of  the 

12  ft      7'  multiplicand  by  the  feet  in  the  multiplier,  be- 

4  ft      3/  ginning  at  the  right  hand.     Thus,  4  times  7' 

£o~ft  —  ^  -       are  28',  equal  to  2  ft.  and  4'.     Set  the  4 

3  ft      1'     9"      under  inches,  and  carry  the  2  feet  to  the  next 

-53  £t'  —  ^  —  ^77      product.     4  times  12  ft.  are  48  ft.  and  2  to 

carry  make  50  ft.     Again,  since  3'=-f\  of  a 

ft.  and  7'=^r  of  a  ft->  3/  into  1'  is  iVt  of  a  ft.  =  21",  or  1'  and  9y 

Write  the  9"  one  place  to  the  right  of  inches,  and  carry  the  1'  to 

the  next  product.     Then  3'  or  -^  of  a  ft.  multiplied  into  12  ft.= 

ff  of  a  ft.,  or  36',  and  1'  to  carry  make  37'  ;  but  37'=3  ft.  and  1'. 

Now  adding  the  partial  products,  the  sum  is  53  ft.  5'  9''. 

Otis.  It  will  be  seen  from  this  operation,  that  feet  multiplied  into  feet,  pro- 
duce feet  ;  feet  into  inches,  produce  inches  ;  inches  into  inches,  produce 
seconds,  &c.  That  is,  the  product  of  any  two  factors  has  as  many  accents  as 
the  factors  themselves  have.  Hence, 

515*  To  find  the  denomination  of  the  product  of  any  two 
factors  in  duodecimals. 

Add  the  indices  of  the  two  factors  together,  and  the  sum  will  be 
the  index  of  their  product. 

Thus,  feet  into  feet,  produce  feet;  feet  into  inches,  produce 
inches  ;  feet  into  seconds,  produce  seconds  ;  feet  into  thirds,  pro- 
duce thirds  ;  £c. 

Inches  into  inches,  produce  seconds  ;  inches  into  seconds,  pro- 
duce thirds  j  inches  into  fourths,  produce  fifths,  (fee. 

Seconds  into  seconds,  produce  fourths  ;  seconds  into  thirds,  pro- 
duce fifths  j  seconds  into  sixths,  produce  eiyhtks,  &c. 

Thirds  into  thirds,  produce  sixths  /  thirds  into  fifths,  produce 
eighths  ;  thirds  into  sevenths,  produce  tenths,  &c. 

Fourths  into  fourths,  produce  eighths  ;  fourths  into  eighths,  pro- 
duce twelfths,  &c. 

Note.  —  The  fooMs  considered  the  unit  and  has  no  index. 


Q.TTEST.— 515.  Hflw  find  the  denomination  of  the  product  in  duodecimals  ?  What  do  feet 
Into  feet  produce  ?  Feet  into  inches?  Feet  into  seconds ?  What  do  inches  into  inchefl 
produce  1  Inches  into  thirds  ?  Inches  into  fourths  ?  Seconds  into  seconds  ?  Seconds 
Into  thirds  ?  Seconds  into  eighths  7  Thirds  inl  i  thirds  ?  Thirds  into  siiths  ? 


336  DUODECIMALS.  [SECT.  XV. 

516*  From  these  illustrations  we  derive  the  following 

RULE  FOR  MULTIPLICATION  OF  DUODECIMALS. 

1.  Place  the  several  terms  of  tlie  multiplier  under  the  corres2>cnd~ 
ing  terms  of  the  multiplicand. 

II.  Multiply  each  term  of  the  multiplicand  by  each  term  of  the 
multiplier  separately,  beginning  with  the  lowest  denomination  in  the 
multiplicand,  and  the  highest  in  tlie  multiplier,  and  write  the  first 
figure  of  each  partial  product  one  place  to  the  right  of  that  of  the 
pr  seeding  product,  under  its  corresponding  denomination.  (Art.  515.) 

III.  Finally,  add  the  several  partial  products  toy  ether,  carrying 
I  for  every  12  both  in  multiplying  and  adding,  and  tJie  sum  will 
be  the  answer  required. 

UBS.  1.  It  is  sometimes  asked  whether  the  inches  in  duodecimals,  are  linear, 
square,  or  cubic.  The  answer  is,  they  are  neither.  An  inch  is  I  twelfth  of  a 
foot.  Hence,  in  measuring  surfaces  an  inch  is  -fa  of  a  square  foot ;  that  is,  a 
surface  1  foot  wide  and  1  inch  long.  In  measuring  solids,  an  inch  denotes  -fV 
of  a  cubic  foot.  In  measuring  lumber,  these  inches  are  commonly  called  car- 
penter's inches. 

2.  Mechanics,  also  surveyors  of  wood  and  lumber,  in  taking  dimensions  of 
their  work,  lumber,  &c.,  often  call  the  inches  a  fractional  part  of  a  foot,  and 
then  find  the  contents  in  feet  and  a.  fraction  of  a  foot.     Sometimes  inches  are 
regarded  as  decimals  of  a  foot. 

3.  We  have  seen  that  one  of  the  factors  in  multiplication,  is  always  to  be 
considered  an  abstract,  number.  (Art.  82.  Obs.  2.)     How  then,  can   feet  be 
multiplied  by  feet,  inches  by  inches,  &c. 

It  should  be  observed,  that  when  one  geometrical  quantity  is  multiplied  by 
another,  some  particular  extent  is  to  be  considered  the  unit.  It  is  immaterial 
what  this  extent  is,  provided  it  remains  the  same  in  the  different  parts  of  the 
same  calculation.  Thus,  if  one  of  the  factors  is  one  fool  and  the  other  half  & 
foot,  the  former  being  12  in.,  and  the  latter  6  in.,  the  product  is  72  in.  Though 
it  would  be  nonsense  to  say  that  a  given  length  is  repeated  as  often  as  another 
is  long,  yet  there  is  no  impropriety  in  saying  that  one  is  repeated  as  many  times 
as  th.ere  are  feet  or  inches  in  another. 

4.  On  the  principles  of  duodecimals,  it  has  been  supposed  that  pounds 
shillings,  pence,  and  farthings  can  be  multiplied  by  pounds,  shillings,  pence, 
and  farthings.     But  it  may  be  asked,  what  ilcnominalion  shillings  multiplied 
by  pence,  or  pence  by  farthings,  will  produce?      It  is   absurd  to  say  that 
2s.  and  6d.  is  repeated  2s.  and  6d.  times. 

QUEST.— 516.  What  is  the  rule  for  multiplication  of  duodecimals  ?  Obs.  What  kind  of 
Inches  are  those  spoken  of  in  me  curing  surfaces  by  duodecimals  ?  In  measuring  solids  1 


ART.  51 6. J  DUODECIMALS.  337 

Ex.  2.  How  many  square  feet  are  there  in  a  piece  of  marble 
0  ft.  7  in.  2"  long,  and  3  ft.  4  in.  7"  wide  ? 

Note. — It  is  not  absolutely  necessary  to  begin  to  multiply  by  the  highest  de- 
nomination of  the  multiplier,  or  to  place  the  lower  denomination  to  the  right  of 
the  multiplicand.  The  result  will  be  the  same  if  we  begin  with  the  lowest  de- 
nomination of  the  multiplier,  and  place  the  first  figure  of  each  partial  product 
nnder  the  figure  by  which  we  multiply. 

Common  Method.  Second  Method. 

9  ft.     7'     2'  9  ft.     7'    2'' 

3  ft.     4'     V  3  ft.     4'    7-' 


28  ft.     9'     6"  5'     7"        2'" 

3  ft.     2'     4"     8'"  3  ft.     2'     4"        8'" 

5'     7"     2'"  2'"'  28  ft.     9'     6" 


Ans.    32  ft.     5'     5"  10'"  2"".  Ans.  32  ft.     5'     5''       10'"  2"" 

3.  How  many  square  feet  are  there  in  aboard  15  ft.  7  in.  long, 
and  1  ft.  10  in.  wide  ? 

4.  How  many  cubic  feet  in  a  stick  of  timber  15  ft.  3  in.  long, 
2  ft.  4  in.  wide,  and  1  ft.  8  in.  thick? 

6.  How  many  cubic  feet  in  a  block  of  granite  18  ft.  5  in.  long, 
4  ft.  2.  in.  wide,  and  3  ft.  6  in.  thick  ? 

6.  How  many  square  feet  in  a  stock  of  10  boards,  15  ft.  8  in. 
long,  and  1  ft.  6  in.  wide  ? 

7.  How  many  square  feet  in  a  stock  of  15  boards,  20  ft  3  in. 
long,  and  2  ft.  5  in.  wide  ? 

8.  Multiply  16  ft.  3'  4"  by  6  ft.  5'  8"  10"'. 

9.  Multiply  20  ft.  4'  8"  5'"  by  7  ft.  6'  9  '  4"  . 

10.  Multiply  18  ft.  0'  5'-  10'"  by  4  ft.  8'  7"  9'". 

11.  Multiply  50  ft.  6'  0'   2'"  6""  by  3  ft.  10'  5". 

12.  How  many  cords  in  a  pile  of  wood  50  ft.  6  m.  long,  8  ft. 
8  in.  wide,  and  7  ft.  4  in.  high  ? 

13.  If  a  cistern  is  30  ft.  10  in.  long,  12  ft.  3  in.  wide,  and  10  ft. 
2  in,  deep,  how  many  cubic  feet  will  it  contain  ? 

14.  What  will  it  cost  to  plaster  a  room  20  ft.  6  in.  long,  18  ft. 
wide,  and  10  ft.  high,  at  12£  cts.  per  square  yard  ? 

15.  How  many   bricks  8  in.  long,  4  in.  wide,  and   2  in.  thick, 
will  make  a  wall  50  ft  long,  10  ft.  high,  and  2  ft.  6  hi.  thick  2 


338  EQUATION  [SECT.  XVI 

SECTION    XVI. 
EQUATION    OF    PAYMENTS. 

ART  517  EQUATION  OF  PAYMENTS  is  the  process  of  finding 
he  equalized  or  average  time  when  two  or  more  payments  due  at 
lifferent  times,  may  be  made  at  once,  without  loss  to  either  party. 

OBS.  The  equalized  or  average  time  for  the  payment  of  several  debts,  due  at 
different  times,  is  often  called  the  mean  time. 

518*  From  principles  already  explained,  it  is  manifest,  when 
the  rate  is  fixed,  the  interest  depends  both  upon  the  principal  and 
the  time.  (Art.  404.)  Thus,  if  a  given  principal  produces  a  cer- 
tain interest  in  a  given  time, 

Double  that  principal  will  produce  twice  that  interest ; 
Half  that  principal  will  produce  half  that  interest ;  &c. 
In  double  that  time  the  same  principal  will  produce  twice  that 
interest ; 

In  lialf  that  time,  half  that  interest ;  &c. 

519.  Hence,  it  is  evident  that  any  given  principal  will  pro- 
duce the  same  interest  in  any  given  time,  as 

One  half  that  principal  will  produce  in  double  that  time ; 
One  third  that  principal  will  "  "  thrice  that  time ; 
Twice  that  principal  will  "  "  half  that  time ; 

Thrice  that  principal  will  "        "  a  third  of  that  time ;  &c. 

For  example,  at  any  given  per  cent. 

The  int.  of  $2  for  1  year,  is  the  same  as  the  int  of  $1  for  2  yrs. ; 
The  int.  of  $3  for  1  year,    "         "  $1  for  3  yrs. ;    (fee, 

The  int.  of  $4  for  1  mo.      "         "  "        $1  for  4  mos. ; 

The  int.  of  $5  for  1  mo.      "         "  "       $1  for  5  mos. ;  <fec 


QUEST.— 517.  What  is  equation  of  payments  ?  Obs.  What  is  the  average  time  for  tlw 
payment  of  several  debts  sometimes  called?  518.  When  the  ra'«  is  fixed,  upon  wha 
does  the  interest  depend  t 


ARTS.  517-521.]  OF  PAYMENTS.  339 

5  2O.  The  interest,  therefore,  of  any  given  principal  for  1  yeai, 
or  1  month,  d'C.,  is  the  same,  as  the  interest  of  1  dollar  for  as  many 
years,  or  months,  as  there  are  dollars  in  the  given  principal. 

Ex.  1.  Suppose  you  owe  a  man  $15,  and  are  to  pay  him  $5  in 
10  months,  and  $10  in  4  months,  at  what  time  may  both  pay- 
ments be  made  without  loss  to  either  party  ? 

Analysis. — Since  the  interest  of  $5  for  1  month  is  the  same  as 
the  interest  of  $1  for  5  months,  (Art.  519,)  the  interest  of  85  for 
10  months  must  be  equal  to  the  interest  of  $1  for  10  times  5 
months.  And  5  mo.  X  10 —  50  mo.  In  like  manner  the  interest 
of  $10  for  4  months  is  equal  to  the  interest  of  $1  for  4  times  10 
months;  and  10  mo.  X  4  =  40  months.  Now  50  months  added  to 
40  months  make  90  months ;  that  is,  you  are  entitled  to  the  use 
of  $1  for  90  months.  But  $1  is  -fa  of  $15",  consequently  you  aro 
entitled  to  the  use  of  $15,  -^  of  90  months,  and  90-f-15  =  6. 

Ans.  6  months 
Proof. 

The  interest  of  $5    at  6  per  cent,  for  10  mo.  is  $5    X-05  =  $.25 
The  interest  of  $10       "       "  "      4  mo.  is  $10X.02=   .20 

Sum  of  both     $^45~ 

The  interest  of  $15  at  6  per  cent,  for  6  mo.  is  15X-03=$.45. 

521.  From  these  principles  we  derive  the  following  general 

RULE  FOR  EQUATION  OF  PAYMENTS. 

First  multiply  each  debt  by  the  time  before  it  becomes  due  y  then 
divide  the  sum  of  the  products  thus  obtained  by  the,  sum  of  the  debts,, 
and  tJie  quotient  will  be  the  average  time  required. 

OBS.  1.  If  one  of  the  debts  is  paid  down,  its  product  will  be  nothing ;  biu 
in  finding  the  sum  of  the  debts,  this  payment  must  be  added  with  the  others. 

2.  When  there  are  months  a^d  days,  the  months  must  be  reduced  to  days, 
or  the  days  to  the  fractional  part  of  a  month. 

3.  This  rule  is  based  upon  the  supposition  that  discount  and  interest  paid  in 
Advance  are  equal.      But  this  is  not  exactly  true;    consequently,  the  rule, 
though  in  general  use,  is  not  strictly  accurate.  (Art.  432.  Obs.  1.) 


GUEST.— 521.  What  is  the  rule  for  equation  of  payments  ? 
15* 


340  PARTNERSHIP.  [SECT.   XV7!. 

2.  If  you  owe  a  man  $60,  payable  in  4  months,  8120  payable 
in  6  months,  and  $180  payable  in  3  months,  at  what  time  may 
you  justly  pay  the  whole  at  once  ? 

Operation. 

$  60X4   —  $240,  the  same  as  $1  for  240  months.  (Art.  520.) 
$120X6   =  $720,    "      "      "  $1  for  720 
$180X3   =   $540,    "       "      "  $1  for  540 
$360  debts.  $1500,  sum  of  products. 
Now  1 500-7- 360=4i  months.  Ans. 

3.  A  merchant  bought  one  lot  of  goods  for  $1000  on  5  months  ; 
another  for  $1000  on  4  months  ;  another  for  $1500  on  8  months; 
what  is  the  average  time  of  all  the  payments  ? 

4.  If  a  man  has  one  debt  of  $150,  due  in  3  months ;  anothei 
of  $200,  due  in  4£  months  ;  another  of  $500,  due  in  7£  months : 
what  is  the  average  time  of  the  whole  ? 

O 

5.  A  man  bought  a  house  for  $3500,  and  agreed  to  pay  $500 
down,  and  the  balance  in   6  equal  annual  instalments :  at  what 
time  may  he  pay  the  whole  ? 

6.  If  you  owe  one  bill  of  $175,  due  in  30  days  ;  another  of  $81, 
due  in  60  days ;  another  of  $120,  due  in  65  days,  and  another  of 
$200,  due  in  90  days  :  when  may  you  pay  the  whole  at  once  ? 

PARTNERSHIP. 

522.  PARTNERSHIP  is  the  associating  of  two  or  more  individ- 
uals together  for  the  transaction  of  business.  (Art.  464.)  The  per- 
sons thus  associated  are  called  partners;  and  the  association 
itself,  a  company  or  firm. 

The  money  employed  is  called  the  capital  or  stock ;  and  the 
profit  or  loss  to  be  shared  among  the  partners,  the  dividend. 

CASE  I. —  When  stock  is  employed  an  equal  length  of  time. 

Ex.  1.  A  and  B  formed  a  partnership ;  A  furnished  $600  cap- 
ital, and  B  $900  •,  they  gained  $300  :  what  was  each  partner's 
share  of  the  gain  ? 

QCEST.—  ">22.  What  Is  partnership  ?  What  are  the  persons  thus  associated  called  t 
What  is  the  association  itself  called  7  What  is  the  money  employed  called  ?  What  th« 
profit  or  loss  ? 


ARTS.  522,  523.]  PARTNERSHIP.  341 

Analysis. — Since  the  whole  stock  is  $600-{-$900  —  $1500,  A'a 
part  of  it  was  -&uV— £>  and  B's  part  was  iWo=f-  Now  since 
A  put  in  -f  of  the  stock,  he  must  have  f  of  the  gain ;  and  $300 
Xf=$120.  For  the  same  reason  B  must  have  £  of  the  gain; 
and  $300Xf=$180. 

Or,  we  may  reason  thus  :  As  the  whole  stock  is  to  the  whole 
gain  or  loss,  so  is  each  man's  particular  stock  to  his  share  of 
the  gain  or  loss. 

That  is,   $1500  :  $300  :  :  $600  :  A's  gain  ;  or  $120, 
And        $1500  :  $300  :  :  $900  :  B's  gain  ;  or  $180. 

PROOF. — $120+ $180=$300,  the  whole  gain.  (Art.  21.  Ax.  11.) 

523.  Hence,  to  find  each  partner's  share  of  the  gain  or  loss, 
when  the  stock  of  each  is  employed  for  the  same  time. 

Multiply  each  man's  stock  by  the  whole  gain  or  loss  ;  divide  the 
product  by  the  whole  stock,  and  the  quotient  will  be  his  sJiare  of  the 
gain  or  loss. 

Or,  make  each  mans  stock  the  numerator,  and  the  whole  stock 
the  denominator  of  a  common  fraction  ;  multiply  the  gain  or  loss 
by  the  fraction  which  expresses  each  mans  share  of  the  stock,  and 
the  product  will  be  his  share  of  the  gain  or  loss. 

PROOF. — Add  the  several  shares  of  the  gain  or  loss  together,  and 
if  the  sum  is  equal  to  the  whole  gain  or  loss,  tJie  work  is  right. 
(Art.  21.  Ax.  11.) 

OBS.  1.  The  preceding  case  is  often  called  Singh  Fellowship.  But  since  a 
pnrf.nerskfp  is  necessarily  composed  of  (wo  or  more  individuals,  it  is  somewhat 
difficult  to  see  the  propriety  of  calling  it  single. 

2.  This  rule  is  applicable  to  questions  in  Bankruptcy,  and  all  other  opera- 
tions in  which  there  is  to  be  a  division  of  property  in  specified  proportions. 
(Arts,  405,  466.) 

2.  A,  B,  and  C  formed  a  partnership  ;  A  put  in  $1200  of  the 
capital,  B  $1600,  and  C  $2000  ;  they  gained  $960  :  what  was 
each  man's  share  of  the  gain  ? 

QrnsT.— 523.  How  is  each  man's  share  of  the  gain  or  loss  found,  when  the  stock  <rf 
Bach  is  employed  for  the  same  time  1  How  is  the  opera  tio  a  proved?  Obs.  VVhstt  is  U 
sometimes  called  ?  To  what  is  this  rule  applicable  1 


342  GENERAL  [SECT     XVI 

3.  A,  B,  and  C  entered  into  partnership ;  A  furnished  $2350, 
B  $3200,  and  C  $1820  ;  they  lost  $860  :  what  ^ras  eacl   man's 
share  of  the  loss  ? 

4.  A  bankrupt  owes  A  $2400,  B  $4600,  C  $6800,  and  D  $9000 ; 
his  whole  effects  are  worth  $11200 :  how  much  will  each  creditor 
receive  ? 

5.  A,  B,  C,  and  D,  engaged  in  an  adventure  ;  A  put  in  $170, 
B  $160,  0  $140,  and  D  $130  ;  they  made  $3000  :  what  was  each 
man's  share  ? 

CASE  II. —  When  the  stocks  are  employed  unequal  lengths  of  time. 

6.  A  and  B  formed  a  partnership  ;  A  put  in  $000  for  4  months, 
and  B  put  in  $400  for  12  months;  they  gained  $763  :  what  was 
each  man's  share  of  the  gain? 

Note. — It  is  obvious  that  the  gain  of  each  depends  both  upon  the  Capital  he 
furnished,  and  the  lime  it  was  employed.  (Art.  518.) 

Analysis. — Since  A's  capital  $900,  was  employed  4  months, 
his  share  of  the  gain  is  the  same  as  if  he  had  put  in  $3600  for  1 
month;    (Art.   519;)    for  $900X4=$3600.      Also    B's  capital 
$400,  being  employed  12   months,  his  share  of  the  gain  is  the 
same  as  if  he  had  put  in  $4800  for  1  month;  for  $400X12  = 
64800.     The  sum  of  $3600  and  $4800  is  $8400.     Therefore, 
A's  share  of  the  gain  must  be  f|^=f . 
B's      "         "          "         "        if££=4-. 
Now  $763 Xf =$327,  A's  share. 
And  $7.63 xf=$436,  B's  share.     Hence, 

524*  To  find  each  partner's  share  of  the  gain  or  loss,  when 
vae  stock  of  each  is  employed  'unequal  lengths  of  time. 

Multiply  each  partner's  stock  by  the  time  it  is  employed ;  make 
each  mans  product  the  numerator,  and  the  sum  of  the  products  the 
denominator  of  a  common  fraction  ;  then  multiply  the  whole  gain 
or  loss  by  each  man's  fractional  share  of  the  stock,  and  the  product 
vill  be  his  share  of  the  gain  or  loss. 

OBS.  This  case  is  often  called  Compound  or  Double  Fellowship. 

QUEST. — 524.  When  the  stock  of  each  partner  is  employed  uneqnr  I  lengths  »f  time, 
how  is  each  man's  share  found  1  06*.  What  is  this  case  sometimes  ctl'.ed  ? 


ARTS.  524-527.]  AVERAGE.  343 

7.  The  firm   of  X,  Y,  and   Z  lost  $4500 ;  X  had  13200  em- 
ployed for  6  months,  Y  -$2400  for  7  months,  and  Z  $1800  for  9 
months  :  what  was  each  partner's  loss  ? 

8.  A,  B,  and  C  hired  a  pasture  for  $60;    A  put  in  15  oxen 
for  20  days,  B  17  oxen  for  16^  days,  and  C  22  oxen  for  10  days  : 
what  r?nt  ought  each  man  to  pay  ? 

9.  In  a  certain  adventure  A  put  in  $12000  for  4  months,  then 
adding  $8000  he  continued  the  whole  2  months  longer ;    B  put 

n  $25000,  and  after  3  months  took  out  $10000,  and  continued  the 
rest  for  3  months  longer ;  C  put  in  $35000  for  2  months,  then 
withdrawing  f  of  his  stock,  continued  the  remainder  4  months 
longer ;  they  gained  $15000  :  what  was  the  share  of  each  ? 

GENERAL   AVERAGE. 

525*  The  term  General  Average,  in  commerce,  signifies  the 
apportionment  of  certain  losses  among  the  different  interests  con- 
cerned, when  a  part  of  the  cargo,  furniture,  &c.,  of  a  ship  has 
been  voluntarily  sacrificed  to  preserve  the  rest.  (Art.  466.) 

The  property  thus  sacrificed  is  called  the  jettison. 

520.  Losses  thus  incurred  are  charged  to  the  ship,  the  cajffor 
and  the  freir/ht,  pro  rata  j  or  according  to  the  value  of  each. 

The  contributory  interests  are  to  be  freed  from  all  charges  upon 
them  before  the  average  is  made. 

OBS.  1.  In  estimating  the  freight,  in  New  York,  one-half,  but  in  most  ports 
one-third  is  deducted  from  the  gross  amount,  for  seamen' 's  wages,  pilotage,  and 
other  small  charges. 

2.  In  the  valuation  of  masts,  spars,  cables,  rigging,  &c.,  of  the  ship,  it  is  cus- 
tomary to  deduct  a  third  from  the  cost  of  replacing  them;  thus  calling  the 
old,  two-thirds  the  value  of  the  new,  in  making  the  average. 

3.  The  cargo  is  valued  at  the  price  it  would  bring  at  its  destined  port,  after 
the  storage  and  other  necessary  charges  are  deducted.     The  property  sacri- 
ficed must  be  taken  into  the  account  as  well  as  that  which  is  saved. 

527*  General  Average  may  be  calculated  both  by  Analysis 
and  Partnership.  (Arts.  464,  522.) 

OBS.  1.  Losses  arising  from  the  ordinary  wear  and  tear,  or  from  a  sacrifice 
made  for  the  safety  of  the  ship  only,  or  a  particular  part  of  the  cargo,  must 
be  borne  by  tfr  3  individuals  who  own  the  property  lost,  and  hot  by  general 
Average 


344 


EXCHANGE    OP 


[SECT.  XVI. 


2.  General  average  is  not  allowed,  unless  the  peril  was  imminent •$,  and  the 
sacrifice  indispensable  for  the  safety  of  the  ship  and  crew. 

10.  The  ship  Minerva  from  London  to  New  York,  had  on  board 
a  cargo  valued  at  $75000,  of  which  A  owned  $30000  ;  B  $27000  ; 
and  C  "$18000;  the  gross  amount  of  freight  and  passage  money 
was  $11040.  The  ship  was  worth  $40000,  and  the  owner  paid 
$520  for  insurance  on  her.  Being  overtaken  by  a  severe  tempest, 
the  master  threw  $18000  worth  of  A's  goods  overboard,  and  cut 
away  her  mainmast  and  anchors ;  finally,  he  brought  her  into 
port,  where  it  cost  $2796.75  to  repair  the  injury:  what  was  the 
loss  of  each  owner  of  the  ship  and  cargo  ? 


Operation. 
Ship  valued  at          ...          $40000.00 
Less  premium  for  insurance          .             520.00 

$39480.00 

Cargo  worth       .... 
Freight  and  passage  money         .       $11040.00 
Less  one-half  for  wages  of  crew             5520.00 

75000.00 
5520.00 

Amount  of  contributory  interests 

Goods  thrown  overboard  valued  at 
Cost  new  masts,  spars,  &c.           .         $2796.75 
Less  one  third  for  wear  of  old      .             932.25 
Commissions  on  repairs 
Port  duties  and  incidentals 

$120000.00 
$18000.00 

1864.50 
15.13 
120.37 

Amount  of  loss 

Now  $20000X30000-$120000=$5000,  loss 
$20000X27000-$120000=$45«0 
$20000X18000—  $120000=$3000 
$20000X39480—  $120000=^6580 
$20000X    5520—  $120000  =  8  920 

$20000.00 

of  A. 
B. 
C. 

Ship. 
Freight 

PROOF.  —  Whole  loss  (Ax.  11.)  $20000,  the 

same  as  above. 

Note. — We  may  also  find  what  percent,  the  loss  is;  then  multiply  each 
contributory  interest  by  the  per  cent.  Thus,  since  $120000  lose  &.0000,  $1  will 
|osc  T^_JL__  Of  $20000;  and  20000-?-$! 20000 =.  1 6f  ;  that  is,  the  loss  is  Ibf 
per  cent  Now  $30000 X-16f =$5000,  A's  share  of  the  loss.  The  loss  of  the 
others  may  be  found  in  a  similar  manner. 


ARTS.  528-532.]  CURRENCIES.  345 

EXCHANGE  OP  CURRENCIES. 

528.  The  term  currency,  signifies  money,  or  the  dnmlating 
medium  of  trade. 

5  2O.  The  intrinsic  value  of  the  coins  of  different  nations,  de- 
pends upon  their  weight  and  the  purity  of  pie  metal  of  which  they 
are  made.  (Art.  245.  Obs.) 

OBS.  For  the  present  standard  weight  and  purity  of  the  coins  of  the  United 
States,  see  Arts.  245,  246.  For  that  of  British  coin,  see  Art.  248.  Obs. 

53O«  The  relative  value  of  foreign  coins  is  determined  by  the 
laws  of  the  country  and  commercial  usage. 

OBS,  The  legal  value  of  a  pound  sterling  in  this  country  has  been  different 
at  different  times.  By  act  of  Congress,  1799.  it  was  fixed  at  $4.44 f-.  In  1832 
its  value  was  raised  by  the  same  authority  to  $4.80;  and  in  1842,  to  $4.84. 

531*  The  process  of  changing  money  from  the  denominations 
of  one  country  to  its  equivalent  value  in  the  denominations  of  an- 
other country,  is  called  Exchange  of  Currencies. 

CASE  I. — Reduction  of  Sterling  to  Federal  Money. 
Ex.  1.  Change  £60  sterling  to  Federal  money. 
Solution. — Since  £l  is  worth  $4.84,  £60  are  worth  60  times  as 
much,  and  $4.84X60=$290.40.  Ans. 

2    Change  £8,  7s.  6d.  to  Federal  money. 

Operation.  We  first  reduce  the  7s.  6d.  to  the  decimal 

$4.84  of  a  pound  ;  (Art.  346  ;)  then  multiply  $4.84, 

8.375  and  £8.375  together,  and  point  off  the  prod- 

$40.535  Ans.  uct  as  in  multiplication  of  decimals.     Hence, 

53.2*  To  reduce  Sterling  to  Federal  Money. 

Multiply  the  legal  value  of  one  pound,  $4.84,  by  the  given  num- 
ber of  pounds,  point  off  the  product  as  in  multiplication  of  deci- 
mals, and  it  will  be  the  answer  required.  (Art.  324.) 

If  the  example  contains  shillings,  pence,  and  farthincs,  they  must 
be  reduced  to  the  decimal  of  a  pound. 

QUEST. — 528.  What  is  meant  by  currency  ?  529.  On  what  does  the  ir  trlnsic  val:ie  of 
the  coins  of  different  countries  depend  ?  530.  How  is  the  relative  value  of  foreign  coiiu 
determined?  Obs.  What  is  the  value  of  a  pound  sterling?  531.  What  is  me  tnt  by  ei 
change  of  currencies  1  532.  How  is  Sterling  money  reduced  to  Federal  ? 


346  EXCHANGE    OP  [SECT.  XVI. 

OBS.  1.  The  reason  of  this  rule  is  obvious  from  the  principle  that  £5  are 
worth  5  times  as  much  as  £1,  &c. 

2,  The  rule  usually  given  for  reducing  Sterling  to  Federal  Money,  is  to  re- 
duce the  shillings,  pence,  and  farthings  to  the  decimal  of  a  pound,  and  placing 
it  on  the  right  of  the  given  pounds,  divide  the  whole  sn-m  by  -fa.  This  rule  is 
based  on  the  law  of  1799,  which  fixed  the  value  of  a  pound  at  $4.44$-,  and 
that  rf  a  dollar  at  4s.  6d.  *  But  $4.44^  is  9  per  cent,  of  itself,  or  40  cents  less 
than  $4.84,  which  is  the  present  legal  value  -of  a  pound  ;  consequently,  the 
result  or  answer  obtained  by  it,  must  be  9  per  cent,  too  small.  A  dollar  is  now 
«qual  to  49.6d.  very  nearly,  instead  of  54d.  as  formerly. 

533*  From  the  preceding  rule  it  is  plain  that  Guineas,  Francs 
Doubloons,  and  sin  foreign  coins,  may  be  reduced  to  Federal  money 
by  multiplying  the  legal  value  of  one  by  the  given  number. 

Change  the  following  sums  of  Sterling  to  Federal  money: 

3.  £850,  10s.  8.  £1000,  4s.  6d.  13.  £50173,  12s.  6ld. 

4.  £175,  15s.  9.  £1600,  8s.  7|d.         14.  £53262,  13s.  8£d. 
6.  £85,  13s.  6d.       10.  £12531,  10s.  4£d.     15.  £76387,  15s.  7|d. 

6.  £200,  7s.  6d.       11.  £43116,  9s.  lOd.       16.  £58762,  18s.  9|d. 

7.  £421,  16s.  4d.     12.  £68318,  10s.  3|d.     17.  £1000000. 

CASE  II. — Reduction  of  Federal  to  Sterling  Money. 

18.  Change  $40.535  to  sterling  money. 

Solution. — Since  $4.84  are  worth  £l,  $40.535  are  worth  as 
many  pounds  as  $4.84  are  contained  times  in  $40.535 ;  and 
$40.535-h$4. 84=8. 375  ;  that  is  £8.375.  Now  reducing  the  de- 

7  O 

cimal  .375  to  shillings  and  pence,  (Art.  348,)  we  have  £8,  7s.  6d. 
for  the  answer.     Hence, 

534*  To  reduce  Federal  to  Sterling  money. 

Divide  the  given  sum  by  $4.84,  (the  value  of  £l,)  and  point  off 
t\e  quotient  as  in  division  of  decimals.  Tlie  figures  on  tlie  left 
Jiand  of  the  decimal  point  will  be  pounds  ;  those  on  the  right,  deci- 
mals of  a  pound,  which  must  be  reduced  to  shillings,  pence,  and 
farthings.  (Art.  348.) 

OBS.  Federal  money  may  be  reduced  to  Guineas,  Francs,  or  any  foreign 
coin,  by  dividing  the  given  sum  by  the  value  of  one  guinea,  one  franc,  &c, 

QUEST  —  Obs.  How  may  foreign  coins  be  reduced  lo  Federal  inunei        S34.  Uow  it  Fed 

eral  money  reduced  to  Sterling? 


ARTS.  533-536.]  CURRENCIES.  347 

Change  the  following  sums  of  Federal  to  Sterling  money : 

19.  $'396.88.  23.  $2160.50.  27.  $25265, 

20.  435.60.  24.        975.66.  28.     41470. 

21.  876.25.  25.     4275.10.  29.     50263. 

22.  1265.33,  26.     5300.75.  30.      100000. 

535*  Previous  to  the  adoption  of  Federal  money  in  17 86, 
accounts  in  the  United  States  were  kept  in  pounds,  shillings, 
K'lice,  and  farthings. 

OBS.  At  the  time  Ft  b:al  money  was  adopted,  the  colonial  currency  or  bills 
of  credit  issued  by  the  colonies,  had  more  or  less  depreciated  in  value :  that  is, 
a  colonial  pound  was  worth  less  than  a  pound  Sterling ;  a  colonial  shilling, 
than  a  shilling  Sterling,  &c.  This  depreciation  being  greater  in  some  col- 
onies than  in  others,  gave  rise  to  the  different  values  of  the  State  currencies. 

In  N.  K.  cur.,  Va.,  Ky.,  Tenn.,  la.,  111.,  Miss.,  Missou.,       6s.  or  £-^.=$1. 

In  N.  Y.  cur.,  N.  C.,  Ohio,  and  Mich.,  -      8s.  or  £-£=$!. 

In  Penn.  cur.,  New  Jer.,  Del.,  and  Md.,  7s.  6d.  (7£s.)  or  £f=$l. 

In  Georgia  cur.,  and  South  Carolina,  4s.  8d.  (4|s.)  or  £-j^ =$1. 

In  Canada  cur.,  and  Nova  Scotia,          -  -        5s.  or  £^-=$1. 

Ala.,  La.,  Ark.,  and  Florida  use  Federal  Money  exclusively. 

CASE  III. — Reduction  of  federal  Money  to  State  currencies. 

31.  Reduce  $63.25  to  New  England  currency. 

Solution. — Since  $1  contains  6s.  N.  E.  cur.,  $63.25  contains 
63.25  times  as  many;  and  6s.X63.25  =  379.50s.  Now  379-r20 
=£18,  19s.,  and  .5s.Xl2  =  6d.  (Art.  348.)  Ans.  £18.  19s.  6d. 

536*  Hence,  to  reduce  Federal  money  to  State  currencies. 

Multiply  the  given  sum  by  the  number  of  shillings  which,  in  the 
required  currency,  make  $1,  and  the  product  will  be  the  answer  in 
shillings,  and  decimals  of  a  shilling.  The  shillings  s/tould  be  re~ 
duced  to  pounds,  and  the  decimals  to  pence  and  farthings,  (Art.  348.) 

32.  Reduce  $450  to  New  England  currency. 

33.  Reduce  $567.5C  to  New  York  currency. 

34.  Reduce  $840.10  to  Pennsylvania  currency. 

35.  Reduce  $1500  to  Canada  currency. 

QUEST — 535.  Previous  to  the  adoption  of  Federal  money,  in  what  were  accounts  kept) 
136.  How  is  Fede/M  monev  reduced  to  the  State  currencies  ? 


348  FOREIGN    MONEYS  [SECT.   XVI 

CASE  IV. — Reduction  of  State  currencies  to  Federal  Money. 

36.  Reduce  £23,  12s.  6d.  N.  E.  currency,  to  Federal  money. 
Solution.— £23,  12s.  6d.  =  472.5s.  (Art.  348.)     Now  since  6s. 

N.  E.  cur.  make  $1,  472.5s.  will  make  as  many  dollars  as  6s.  is  con- 
tained times  in  472.5s. ;  and  472.5s.-H6s.  =  78.75.     Ans.  $78.75. 

537.  Hence,  to  reduce  State  currencies  to  Federal  money. 

Reduce  the  pounds  to  shillings,  and  the  given  pence  and  farthing* 
to  the  decimal  of  a  shilling  ;  then  divide  this  sum  by  the  number  of 
shillings  which,  in  the  given  currency,  make  $1,  and  tlie  quotient 
will  be  the  answer  in  dollars  and  cents. 

OBS.  One  state  currency  may  be  reduced  to  another  by  first  reducing  the 
given  currency  to  Federal  money,  then  to  the  currency  required. 

37.  Reduce  £160,  5s.  N.  E.  currency,  to  Federal  money. 

38.  Reduce  £245,  13s.  6d.  N.  Y.  currency,  to  Federal  money. 

39.  Reduce  £369,  15s,  7£d.  Penn.  currency,  to  Federal  money. 

40.  Reduce  £1800,  Georgia  currency,  to  Federal  money. 

41.  Reduce  £5000,  Canada  currency,  to  Federal  money. 

FOREIGN   COINS   AND   MONEYS   OF  ACCOUNT. 

538*  The  denominations  of  money,  in  which  the  laws  of  a 
country  require  accounts  to  be  kept,  are  called  Moneys  of  account. 
They  are  generally  represented  by  a  coin  of  the  same  name ; 
sometimes,  however,  they  are  merely  nominal,  like  mills  in  Fede- 
ral money.  (Art.  245.) 

539»  Foreign  Moneys  of  Account,  with  the  par  value  of  the 
unit  established  by  commercial  usage,  expressed  in  Federal  Money* 

Austria. — 60  kreutzers=:l  florin ;  1  florin,  (silver)  is  equal  to  $0.485 

Belgium. — 100  cents=l  guilder  or  florin;   I  guilder,  (silver)  .40 

The  coinage  of  Belgium  in  1832,  was  made  similar  to  that  of  France, 

Bencoolen. — 8  satellers— I  soocoo;  4  soocoos=l  dollar  or  rial,           -  1.10 

Brazil.— 1000  rees=l  milree=&828.     The  silver  coin,  1200  rees  394 

Bremen. — 5  schwares=l  grote;  72  grotes=l  rix  dollar,  (silver)  .787 

British  India. — 12  pice— 1  anna;  10  annasr=l  Co.  rupee,  (silver)  .445 

The  current  (silver)  rupee  of  Bengal,  Bombay  and  Madras,  is  worth  .444 


QUEST. — 537.  How  are  the  several  State  currencies  reduced  to  Federal  Money? 
*  M'CulIoch's  Commercial  Dictionary ;  Kelly's  Universal  Cambist. 


ARTS.  537-539. J  OF  ACCOUNT.  349 

Buenos  Ayres. — 8  rials  —  1  dollar  currency,  (fluctm  ting)       -  -      $0.93 

Canton, — 10  cash  — 1  candarine;   10 can.  =  1  mace;   10  mace=l  tael         1.48 
The  casn,  which  is  made  of  copper  and  lead,  is  said  to  be  the  only 

money  coined  in  China. 

Cape  of  Good  Hope. — G  stivers— 1  schilling;  Sschillings^l  rix  dollar          .313 
Ceylon. — 4  pice=l  fanam;  12  fan ams=l  rix  dollar  -  -  .40 

Cuba.— 8  rials  plate=l  dollar;  1  dollar        -  -         1.00 

Colombia* — 8  rials=l  dollar;   1  dollar,  (variable)  mean  value         -         1.00 
Chili.—^  rials^l  dollar;  1  dollar,  (silver)  -         1.00 

Denmark. — 12  pfenings— 1  skilling,   16  skillings:=l  marc;  6marcs= 

1  r.gsbank  or  rix  dollar,  (silver)      -  -  .52 

Egypt. — 3  aspers=l  para ;  40  paras— 1  piastre,  (silver)        -  -  .048 

Fiance  and  Great  Britain. — See  Tables.  (Arts.  -247,  272.) 
Greece. — 100  lepta=l  drachme  ;   1  drachma,  (silver)  -  .166 

Holland. — 100  cents=l  florin  or  guilder;  1  florin,  (silver)  .40 

Hamburg. — 12  pfenings^l  schilling  or  sol;    16  schillings=l  marc 
Lubs;  3  rnarcs^l  rix  dollar.     The  current  marc,  (silver)— $-28; 

marc  banco  -------  .35 

The   term   Lubs,  signifies  money  of  Lubec.     The  marc  currency 

is  the  common  coin ;  the  marc  banco  is  based  upon  certificates 

of  deposit  of  bullion  and  jewelry  in  the  bank  of  Hamburg. 
Invoices  and  accounts  are  sometimes  made  out  in  pounds,  schillings, 

and  pence,  Flemish,  whose  subdivisions  are  like  sterling  money ; 

the  pound  Flemish=7£  marcs  banco. 

Japan. — 10  candarines=il  mace;  10  mace  =  l  tael  -  .75 

Java.— 100  cents —1  florin;  1  florin,  as  in  Netherlands         -  -  .40 

Also  5  doits=l  stiver;    2  stivers^  1  dubbel;    3  dub.  =  l  schilling; 

4  schillings^  1  florin        -  -  -  -  .40 

Malta. — 20  granit=l  taro;  12tari  — 1  scudo;  2j  scudi^l  pezza  1.00 

Mauritius. — In  public  accounts  100  cents^rl  dollar  -  -  .968 

In  mercantile  accounts  20  sols=l  livre;  10  livres=l  dollar. 
Manilla. — 34  maravedis— 1  rial;  8  rials =1  dollar,  (Spanish)  -         1.00 

Milan.— 12denari  —  I  soldo;  20  soldi=l  lirat  -  -          .20 

Mexico. — 8  rials— 1  dollar;  1  dollar  -         1.00 

Monte  Video.— 100  centesimos=l  rial;  8rials=l  dollar      -  .833 

Naples. — 10  grani=:l  carlino;  10  carlinir=l  ducat,  (silver)  •  .80 

Netherlands. — Accounts  are  kept  throughout  the  kingdorr  in  florins  or 

guilders,  and  cents,  as  adopted  in  1815.  See  Holland. 
New  South,  Wales. — Accounts  are  kept  in  sterling  money. 
Norway. $ — 120  skillings;=l  rix  dollar  specie,  (silver)  -  1.06 

Papal  Slates. — 10  bajocchi  =  l  paolo;    10  paoli^l  scudc  or  crown  1.00 

Pern.— 8  rialsr=l  dollar,  (silver)        -  -  -  -  -         1.00 


*  Venezuela,  New  Grenada,  and  Ecuador. 

t  Grani  is  the  plural  of  grano,  tari  of  taro,  scudi  of  scudr   lire  of  lira,  pezze  of  pezza. 

(  Norway  has  no  national  gold  coin 


850 


FOREIGN    COINS. 


[SECT.  XVI. 


Portugal. — 400  rees=l  cruzado;  1000rees=l  milree  or  crown  -  $1.12 
Prussia. — 12  pfenings=l  grosch,  (silver)  30  groschen^l  thaler  ordol.  .69 
Russia.* — 100  copecksr=l  rouble,  (silver)  -  -  .78 

Sardinia, — 100  centesimal  lira;  1  lira=l  franc,  French  .186 

Sweden. — 12  rundstycks  =  l  skilling;  48  skillings:=l  rix  dol.,  specie  1.06 
Sicit.y.—SQ  grani=l  taro;  30  tari— 1  oncia,  (gold)  -  2.40 

Spain. — 2  maravedis=l   quinto;   16  quintos=rl  rial  of  oW  plate     -          .10 
20  rials  vellon=l  Spanish  dollar  -         1.00 

The  rial  of  old  plate  is  not  a  coin,  but  it  is  the  denomination  in 

which  invoices  and  exchanges  are  generally  computed. 

SI.  Domingo. — 100  centimes =1  dollar;  I  dollar       -  33| 

Tuscany. — 12  denari  di  pezza^l  soldo  di  pezza;  2  soldi  di  pezza=l 

pezza  of  8  rials ;  1  pezza,  (silver)  .90 

Turkey.— -3  aspers— 1  para;  40  paras- 1  piastre,  (fluctuating)        -  .05 

Venice. — 100  centesimal  lira;  1  lira=l  franc,  French        -  .186 

Formerly  accounts  were  kept  in  ducats,  lire,  &c.  12  denari=l 
soldo;  20  soldi=llira  piccola;  6^  lire  piccole=l  ducat  current; 
8  lire  pic.=l  ducat  effective.  The  value  of  the  lira  piccola  is  ,09f 

West.  Indies,  British. — Accounts  are  kept  in  pounds,  shillings,  pence 

and  farthings,  of  the  same  relative  value  as  in  England.     The 

value  of  the  pound  varies  very  much  in  the  different  islands,  and 

is  in  all  cases  less  than  the  pound  sterling. 

5  4O.  The  following  coins  and  moneys  of  account  have  been 
made  current  in  the  United  States,  by  act  of  Congress,  at  the  rates 
annexed.  \ 


Pound  sterling  of  Gt.  Britain,    $4.84 
Pound  of  Canada,  Nova  Scotia, 
Do.  New  Brunswick  and  New- 
foundland,         .         .         .  4.00 
Franc  of  France  and  Belgium,  .186 
Livre  Tournois  of  France,     .  .185 
Florin  of  Netherlands,            .  .40 
Do.  Southern  States  Germany,  .40 
Guilder  of  Netherlands,          .  40 
Real  Vellon  of  Spain,             .  .05 
Do.  Plate  of  Spain,        .         .  .10 
Milree  of  Portugal,         .         .  1.12 
Do.  Azores,            .         .        .  .83^ 
Marc  Banco  of  Hamburg.      .  .35 
Thaler  or  Rix  Dollar,  Prussia, 
and  North.  States  Germany,  .69 


Rix  Dollar  of  Bremen,  .  $0.78f 

Specie  Dollar  of  Denmark,  1.05 

Do.  Sweden  and  Norway,  .  1.06 

Rouble,  silver,  of  Russia,  .  .75 

Florin  of  Austria,           .  .  .485 
Lira    of   Lombardo,    Venetian 

kingdom,             .         .  .  .16 

Lira  of  Tuscany,             .  .  .16 

Do.  of  Sardinia,              .  ,  ,186 

Ducat  of  Naples,            .  .  ,80 

Ounce  of  Sicily,              .  ,  2.40 

Leghorn  Livres,              .  ,  .16 

Tael  of  China,       .        .  .1.48 

Rupee,  Company,          .  .  .445 

Do.  of  British  India,      .  .  .445 

Pagoda  of  India,            .  .  1.84 


*  Previous  to  1840,  accounts  were  kept  in  paper  roubles,  3£  of  which  made  a  silver 
rouble.  t  Laws  of  United  States. 


Amv.  540-543.]  EXCHANGE.  351 

541.  Foreign  gold  and  silver  coins,  at  the  rates  established  by 
the  Custom  Houses  and  commercial  usage* 

Guinea,      English.       (gold)    $5.00      Leghorn  Dollar,  (s.)      $0.90 

Crown,  "  (silver)     1.12    j  Scuda  of  Malta,  (s.)          ,40 


S  hilling  piece, "  (s.)  .23 

Bunk  token,     "  (S.)  .25 

Florin  of  Basle,  (s.)  .41 

Moiclore,  Brazil,  (».)  4.80 


Doubloon,  Mexico,  (#.)  15.60 

Livre  of  Neufchatel,  (>.;  .26* 

Half  Joe,  Portugal,  (g.)  8.53 

Florin,  Prussia,  (s.)  .22f 


Livre  of  Catalonia.  (s.)          ,53j  I  Imperial,  Russia,  (».)        7.83 


Florence  Livre,  (s.)  .15 

Louis  d'or,  French,         (g.)        4.56 
Crown,  "  (s.)        1.05 


Rix  Dollar,  Rhenish,       (.s.)  .60$ 

Rix  Dollar  of  Saxony,     (s.)  .69 


Pistole,  Spanish,  (ST.)  3.97 

40  Francs,        "                (if.)  7.66  !  Rial             "  (s.)  .12J 

5  Francs,       "                 (s.)  93  |  Cross  Pistareen,  (s.)  .16 

Geneva  Livre,                   (s.)  .21  |  Other  Pistareens,  (*.)  .18 

10  Thalers,  German        (g.)  7.80  !  Swiss  Livre,  (s.)  .27 

10  Pauls,  Italy,                 (s.)  .97  \  Crown  of  Tuscany,  (*.)  1.05 

Jamaica  Pound,  nominal,  3.00  |  Turkish  Piastre,  (s.)  .05 

Note. — The  true  method  of  estimating  the  value  of  foreign  coins,  is  by  their 

weight  and  purity. 

EXCHANGE. 

542*  EXCHANGE,  in  commerce,  signifies  the  receiving  or  pay- 
ing of  money  in  one  place,  for  an  equal  sum  in  another,  by  draft 
or  bill  of  Exchange. 

OBS.  1.  A  Bill  of  Exchange  is  a  written  order,  addressed  to  a  person,  di- 
recting him  to  pay  at  a  specified  time,  a  certain  sum  of  money  to  another  per- 
son, or  to  his  order. 

2.  The  person  who  signs  the  bill  is  called  the  drawer  or  maker ;  the  person 
in  whose  favor  it  is  drawn,  the  buyer  or  remitter ;  the  person  on  whom  it  is 
drawn,  the  drawee,  and  after  he  has  accepted  it,  the  accepter ;  the  person  to 
whom  the  money  is  directed  to  be  paid,  the  payee;  and  the  person  who  has 
legal  possession  of  it,  the  holder. 

3.  On  the  reception  of  a  bill  of  exchange,  it  should  be  immediately  pre- 
sented to  the  drawee  ibr  his  acceptance. 

543.  The  acceptance  of  a  bill  or  draft  is  a  promise  to  pay  it 
at  maturity  or  the  specified  time.  The  common  method  of  accept- 

CirEST. — 542.  What  is  meant  by  exchange  ?  Obs.  What  is  a  hill  of  exchange!  Wha 
Is  the  drawer  of  a  bill  ?  The  drawee  1  The  payee  ?  The  holder  1  543.  What  is  meant 
by  the  acceptance  of  a  bill  ?  What  is  the  common  method  of  accepting  a  bill  ? 

*  See  Manual  of  Gold  and  Silver  Coins  by  Eckfeldt  &  Du  Bois  •  Ogden  on  the  Tariff  of 
1846 ;  Taylor's  Gold  and  Silver  Coin  Examiner 


352  EXCHANGE.  [SECT.  XVI, 

ing  a  bill,  is  for  the  drawee  to  write  liis  name  under  the  word 
accepted,  across  the  bill,  either  on  its  face  or  back.  The  drawee 
is  not  responsible  for  its  payment,  until  he  has  accepted  it. 

OBS.  L.  If  the  payee  wishes  to  sell  or  transfer  a  bill  of  exchange,  it  'is  neces- 
sary for  him  to  endorse  it,  or  write  his  name  on  the  back  of  it. 

2.  If  the  endorser  directs  the  bill  to  be  paid  to  a  particular  person,  it  is 
called  a  special  endorsement,  and  the  person  named,  is  called  the  endorsee. 
I'  the  endorser  simply  writes  his  name  upon  the  back  of  the  bill,  the  endorse- 
ment is  said  to  be  blank.  When  the  endorsement  is  blank,  or  when  a  bill  is 
drawn  payable  to  the  bearer,  it  may  be  transferred  from  one  to  another  at 
pleasure,  and  the  drawee  is  bound  to  pay  it  to  the  holder  at  maturity.  If  the 
drawee  or  accepter  of  a  bill  fail  to  pay  it.  the  endorsers  are  responsible  for  it. 

544.  When  acceptance  or  payment  of  a  bill  is  refused,  the 
holder  should  duly  notify  the  endorsers  and  drawer  of  the  fact 
by  a  legal  protest,  otherwise  they  will  not  be  responsible  for  its 
payment. 

OBS.  1.  A  protest,  is  a  formal  declaration  in  writing,  made  by  a  civil  officer 
termed  a  notary  public,  at  the  request  of  the  holder  of  a  bill,  for  its  non-accept- 
ance, or  non-payment. 

2.  When  a  bill  is  returned  protested  for  non-acceptance,  the  drawer  must 
pay  it  immediately,  though  the  specified  time  has  not  arrived,  otherwise  he  is 
liable  to  prosecution. 

3.  The  time  specified  for  the  payment  of  a  bill  is  a  matter  of  agreement  be- 
tween the  parties  at  the  time  it  is  negotiated.     Some  are  payable  at  sight , 
others  in  a  certain  number  of  days  or  months  after  sight,  or  after  date.     When 
payable  after  sight  or  date,  the  day  on  which  they  are  presented  is  not  reck- 
oned.    WThen  the  time  is  expressed  in  months,  they  are  always  understood  to 
mean  calendar  months.     Hence,  if  a  bill  payable  in  one  month  is  dated  the 
25th  of  January,  it  will  be  due  on  the  25th  of  February.     And  if  it  is  dated 

he  28th,  29th,  30th,  or  3lst  of  January,  it  will  be  due  on  the  last  day  of  Feb- 
ruary. It  is  customary  to  allow  three  days  grace  on  bills  ef  exchange. 

545.  Bills  of  exchange  are  usually  divided  into  inland  and 
foreign  bills.     When  the  drawer  and  drawee  both  reside  in  the 
same  country,  they  are  termed  inland  bills  or  drafts  ;  when  they 
reside  in  different  countries,  foreign  bills. 

OBS.  In  negotiating  foreign  bills,  it  is  customary  to  draw  fh/  -ee  of  the  sa-mt 
date  and  amount,  which  are  called  the  First,  Second  and  Third  of  Exchange} 
and  collectively,  a  Set  of  Exchange.  These  are  sent  by  different  ship*  01 

Q.UKST. — 544.-  When  the  acceptance  or  payment  of  a  bill  is  refused,  what  should  be 
d  we  1  Obs.  What  is  a  protest?  545.  How  are  bills  of  exchange  divided  1  Obs  Whatia 
meant  by  a  set  of  exchange  ? 


ARTS.  544-547.]  EXCHANGE.  353 

conveyances,  and  when  the  first  that  arrives,  is  accepted  or  paid,  the  olaer$ 
become  void.  The  object  of  this  arrangement  is  to  avoid  delays,  which  might 
arise  from  accidents,  miscarriage,  &c. 

FORM  OF  A  FOREIGN  BILL  OF  EXCHANGE. 


Exchange  £1000.  BOSTON,  Oct.  3d,  18-17. 

At  ninety  days  sight  of  this  first  of  Exchange?  (the  second  and  third  of 
the  same  date  and  tenor  unpaid,)  pay  George  Lewis,  Esq.,  or  order,  One  Thou- 
sand Pounds  sterling,  with  or  without  farther  advice. 

JOHN  W.  ADAMB. 
To  Messrs.  ROTHSCHILD  &  Co. 
Hrokers,  London. 

FORM  OF  AN  INLAND  BILL  OR  DRAFT. 


$2500.  NEW  YORK,  Sept.  27th,  1847. 

Thirty  days  after  sight,  pay  to  the  order  of  Messrs.  Newman  &  Co., 
Twenty-five  Hundred  Dollars,  value  received,  and  charge  the  same  to 

MACY  &  WOODBURY. 
To  Messrs.   D.  BAKER  &  Co. 

Merchants,  New  Orleans. 

546.  The  term  par  of  exchange,  denotes  the  standard  by 
which  the  comparative  worth  of  the  money  of  different  countries 
is  estimated.  It  is  either  intrinsic  or  commercial. 

The  intrinsic  par  is  the  real  value  of  the  money  of  different 
countries,  determined  by  the  weight  and  purity  of  their  coin. 

The  commercial  par  is  a  nominal  value,  fixed  by  law  or  commer- 
cial usage,  by  which  the  worth  of  the  money  of  different  countries 
is  estimated. 

OBS.  1.  The  intrinsic  par  remains  the  same,  so  long  as  the  standard  coins 
of  each  country  are  of  the  same  metals,  and  of  the  same  weight  and  purity; 
but,  in  case  the  standard  coins  are  of  different  metals,  the  intrinsic  par  must 
vary,  as  the  comparative  values  of  the  metals  vary. 

2.  The  commercial  par  is  conventional,  and  may  at  any  time  be  changed 
by  law  or  custom. 

547»  By  the  term  course  of  exchange  is  meant  the  current 
price  which  is  paid  in  one  place  for  bills  of  a  given  amount  drawn 
on  another  place. 

OBS.  1.  The  course  of  exchange  is  seldom  statimary  or  at  par.     It  varies 

QCKST. — 546.  What  is  meant  by  par  of  exchange  ?     Inlr.nsic  par  7    Cuipmerc  lal  par  1 


354 


EXCHANGE. 


.   XVI 


according  to  the  circumstances  of  trade.  When  the  balance  of  trade  is  against 
a  country,  that  is  when  the  exports  are  less  than  the  imports,  bills  on  the 
foreign  country  will  be  above  par,  for  the  reason  that  there  will  be  a  greater 
demand  tor  them  to  pay  the  balance  due  abroad.  On  the  other  hand,  when 
the  balance  of  trade  is  in  favor  of  a  country,  foreign  bills  will  be  below  pa?', 
for  the  reason  that  fewer  will  be  required. 

2.  It  should  be  remarked  that  the  course  of  exchange  can  never  exceed 
very  much  the  intrinsic  pafvalue;  for  it  is  plain  that  coin  or  bullion  instead  of 
bills  will  be  remitted,  whenever  the  course  of  exchange  is  such  that  the  ex 
ftnse  of  insuring  and  transporting  it  from  the  debtor  to  the  creditor  country 
&»  less  than  the  premium  for  bills,  and  the  exchange  will  soon  sink  to  par. 

548.  Rates  of  exchange  on  Great  Britain  are  commonly  reck- 
oned at  a  certain  per  cent.,  on  the  old  commercial  par,  instead  of 
the  new  par, 

OBS.  1.  According  to  the  old  par,  the  value  of  a  pound  sterling  is  $4.44-|-f 
as  fixed  by  act  of  Congress,  in  1799.  According  to  the  -new  par  it  is  $4.84. 
The  intrinsic  value  of  a  £  ster.,or  sovereign,  according  to  assays  at  the  U.  S. 
nint,  is  $4.81)1.  The  new  par  is  the  value  fixed  by  the  government  in  1842. 
ind  is  used  in  calculating  duties,  when  the  invoice  is  in  sterling  money. 

2.  The  old  par  is  nine  per  cent,  less  than  the  new  par  or  legal  value ;  conse- 
quently the  rate  of  exchange  must  reach  the  nominal  premium  of  9  per  cent 
before  it  is  at  par  according  to  the  new  standard. 

Table  of  Exchange  showing  the.  value  of  £l  Sterling  from  1  to 
to  Yl\  per  cent,  premium  on  the  old  par  of  $4.44-$-. 


Old  Par 

|>4.444!5i  per  ct. 

$4.689:8  per  ct.     $4.800 

9i-  per  ct.  $4.878 

1  per  ct. 

4.489J6 

4.7llJ8i  "           4.811 

10       "        4.889 

2     " 

4.5336i     " 

4.733:8i  "           4.822 

10i    "        4.911 

3     " 

4.578,7 

4.756  8$  "           4.833 

11       "        4.933 

4     " 

4.622  7^     " 

4.767(9  New  Par  4.844 

Hi     "        4956 

4i  " 

4.644  7i     " 

4.77819iperct.     4.856 

12       "        4.978 

5     " 

4.667  7f     " 

4.789J9i  "          4.807 

12i     "        5.000 

Note. — 1.  When  exchange  is  10  per  cent,  advance  or  over,  on  the  old  par, 
it  will  cause  a  shipment  of  specie  to  England ;  for  the  freight,  interest  and  in- 
surance will  not  amount  to  so  much  as  the  premium.  When  the  premium  is 
less  than  9  per  cent.  English  funds  are,  in  reality,  below  their  intrinsic  par. 

2.  The  practice  of  quoting  rates  of  exchange  at  the  old  par,  is  calculated  to 
lead,  persons  unacquainted  with  the  subject  into  serious  mercantile  mistakes, 
and  to  degrade  our  national  currency  by  making  it  appear  to  foreign  nations 
to  be  so  much  below  par. 


A.RTS.  548,  549.]  EXCHANGE.  355 

* 
Ex.  1.  A  merchant  negotiated  a  bill  of  exchange  on  London  for 

£500,  10s.,   at  8  per  cent,  premium  on  the  old  par :  how  much 
did  he  pay  for  the  bill  ? 

Solution. — £500,  10s.  =  £500. 5.   (Art.  346.) 

Now  $4. 44^X500. 5  =  $2224. 44^  at  the  old  par  value. 
Then  $2224.44jX      .08=      177.95:}-  the  premium. 

The  sum  paid  $2402.40  -4w$. 

Or,  the  val.  of  £l  by  table,  $4.80X500.5 =$2402.40.  Ans. 

2.  A  merchant  negotiated  a  bill  on  Liverpool  for  £1000,  at 

1  per  cent,  discount  from  the  new  par :  what  did  he  pay  for  it  ? 

3.  What  will  a  bill  cost,  on  England,  for  £5265,  13s.  6d.,  at 
8-J-  per  cent,  advance  on  the  old  par  ? 

4.  How  much  is  a  bill  worth  on  France  for  1500  francs,  at 

2  per  cent,  above  par,  which  is  $.186  per  franc? 

5.  What  will  a  bill  cost  on  Paris  for  56245  francs,  exchange 
being  5  francs  and  54  centimes  to  the  dollar  ? 

6.  What  cost  a  bill  of  exchange  on  Hamburg  for  2000  marcs 
banco,  at  1  per  cent,  above  par,  which  is  35  cts.  per  marc  ? 

7.  What  cost  a  bill  of  exchange  on  St.  Petersburg  for  2660 
roubles,  at  2  per  cent,  discount,  the  par  being  75  cts.  per  rouble  ? 

8.  What  cost  an  inland  bill  of  exchange  at  Boston,  on  New 
Orleans,  for  $15265.85,  at  1  per  cent,  advance? 

9.  What  cost  a  draft  at  Albany,  on  Mobile,  for  $20260,  at 
2  per  cent,  premium  ? 

10.  What  cost  a  draft  at  St.  Louis,  on  New  York,  for  $35678, 
at  2-^  per  cent,  premium  ? 

ARBITRATION  .OF   EXCHANGE. 

549*  Arbitration  of  Exchange  is  the  method  of  finding  the 
exchange  between  two  countries  through  the  medium  of  that  of 
other  countries. 

OBS,  1.  When  there  is  but  one  intervening  country,  the  operation  is  terme 
simple  arbitration,  when  more  than  one,  it  is  termed  compound  arbitration. 

2.  Problems  in  Arbitration  of  exchange  are  usually  solved  by  conjoined  pro- 
portion. (Art.  511.)     Care  must  be  taken  to  reduce  all  the  quantities  which 
are  of  the  same  kind,  to  the  same  denomination. 
T.H. 


356  ALLIGATION.  [SECT.  XVI. 

Ex.  1 .  If  the  exchange  of  New  York  on  London  is  8  per  cent, 
advance  on  old  par,  or  $4.80  for  £1  sterling,  and  that  of  Amster- 
dam on  London  is  12  florins  for  £l,  what  is  the  arbitrated  ex- 
Change  of  New  York  on  Amsterdam ;  that  is,  how  many  florins 
are  equal  to  $1  U.  S.  ?  Ans.  $1  =  2£  florins. 

2.  A  merchant  in  Baltimore  wishes  to  remit  1200  marcs  banco 
to  Hamburg,  and  the  exchange  of  Baltimore  on  Hamburg  is  35 
cents  for  1  marc.     He  finds  the  exchange  of  Baltimore  on  Paris 
is  18  cents  for  1  franc;  that  of  Paris  on  London,  is  25  francs  for 
£l  sterling;  that  of  London  on  Lisbon,  is  180  pence  for  3  mil- 
rees  ;  that  of  Lisbon  on  Hamburg,  is  5  milrees  for  18  marcs  banco. 
How  much  will  he  gain  by  the  circuitous  exchange  ? 

Ans.  Direct  Ex.  $420  ;  circuitous  Ex.  $375  :  Gain  $45. 

3.  A  man  in  England  owes  a  man  in  Portugal  £420 ;  the  di- 
rect exchange  from  London  to  Lisbon  is*70d.  for  1  milree ;  but 
the  exchange  between  London  and  Amsterdam  is  48  florins  for 
£l  sterling;    between  Amsterdam  and  Paris  it  is  16  florins  for 
3  francs  ;  and  between  Paris  and  Lisbon  it  is  6  francs  for  2  mil 
rees.     Is  it  better  for  the  man  in  Portugal  to  have  a  direct  remit- 
tance from  London  to  Lisbon,  or  a  circuitous  one  through  Amster- 
dam and  Paris  ? 

ALLIGATION. 

55O.  Alligation  is  the  method  of  finding  the  value  of  a  com- 
pound or  mixture  of  articles  of  different  values,  or  of  forming  a 
compound  which  shall  have  a  given  value.  (Art.  467.) 

OBS.  1.  The  term  alligation  is  derived  from  the  Latin  alligo,  which  signifies 
to  bind  or  tie  together.     It  had  its  origin  in  the  manner  of  connecting  the  nura- 
oers  together  by  a  curve  line  in  the  solution  of  a  certain  class  of  examples. 
2.  Alligation  is  usually  divided  into  Medial  and  Alternate.  (Art.  467.  Ob&) 
Note. — For  a  new  method  of  Alligation  Alternate,  see  Key,  p.  72. 

MEDIAL    ALLIGATION. 

551*  Medial  Alligation  is  the  process  of  finding  the  mean 
price  of  a  mixture  of  two  or  more  ingredients  or  articles  of  dif- 
ferent values. 

Note. — The  term  medial  is  derived  from  the  Latin  met  iius,  signifying  a  mean 
or  average. 


ARTS.  550-554]  ALLIGATION.  357 

552*  To  find  the  mean  value  of  a  mixture,  when  the  quantity 
and  the  price  of  each  of  the  ingredients  are  given. 

Divide  the  whole  cost  of  the  ingredients  by  the  whole  quantity 
mixed,  and  the  quotient  will  be  the  mean  price  of  the  mixture. 

PROOF. — Multiply  the  whole  mixture  by  the  mean  price,  and  if 
the  product  is  equal  to  the  whole  cost,  the  work  is  right. 

Ex.  1.  A  grocer  mixed  10  Ibs.  of  tea  worth  5s.  a  pound,  with 
18  Ibs.  worth  3s.  a  pound,  and  20  Ibs.  worth  2s.  a  pound:  what 
is  the  mixture  worth  per  pound  ? 

Solution. —     10  Ibs.  at  5s.  =  50s. 

18  Ibs.  at  3s.  =  54s. 

20  Ibs.  at  2s.=:40s. 

Whole  quantity  48  Ibs.    and    144s.  whole  cost  of  mixture. 
Now  144s.-i-48  =  3.     Am.  3s.  a  pound. 

2.  A  drover  bought  870  lambs  at  75  cts.  apiece,  and  290  sheep 
at  $1.25  apiece :  what  is  the  mean  price  of  the  lot  per  head? 

3.  A  grocer  mixed  12  gals,  of  wine  at  4s.  lOd.  per  gal.,  with 
21  gals,  at  5s.  3d.,  and  29-J-  gals,  at  5s.  8d. :  what  is  a  gallon  of 
the  mixture  worth  ? 

ALTERNATE     \LLIGATION. 

553*  Alternate  Alligation  is  the  process  of  finding  what  quan- 
tity of  any  number  of  ingredients,  whose  prices  are  given,  will 
form  a  mixture  of  a  given  mean  price. 

Note—  The  term  alternate  is  derived  from  the  Latin  aUernatus,  signifying 
by  turns,  and  in  its  present  application,  refers  to  the  connection  of  the  prices 
which  are  less  than  the  mean  price,  with  those  which  are  greater.  Alternate 
alligation  embraces  three  varieties  of  examples 

CASE     I. 

554.  To  find  the  quantity  of  each  ingredient,  when  its  price 
and  that  of  the  required  mixture  are  given. 

I,  Write  the  prices  of  tJie  ingredients  under  each  other,  beginning 
with  tlie  least ;  tl>£n  connect,  with  a  curve  line,  each  price  which  is  less 
than  that  of  the  mixture  with  one  or  more  of  those  that  are  greater , 
and  each  greater  price  with  one  or  more  af  those  that  are  less. 


358  ALLIGATION.  [SbCT.   XVI, 

II.  Write  the  difference  between  tlie  price  of  the  mixture  and  tliat 
of  each  of  the  ingredients  opposite  tlie  price  with  which  they  are 
connected.  If  only  one  difference  stands  against  any  price,  it  will 
denote  the  quantity  to  be  taken  of  that  price  j  but  if^  there  are  more 
tlutn  one,  their  sum  will  be  the  quantity. 

OBS.  It  is  immaterial  in  what  manner  we  select  the  pairs  of  ingredients, 
provided  the  price  of  one  of  the  ingredients  is  less  and  the  other  greater  than 
fchs  mean  price  of  the  mixture  required. 

PROOF.  —  Find  tlie  value  of  all  the  ingredients  at  their  given 
prices  ;  if  this  is  equal  to  the  value  of  the  whole  mixture  at  tlie 
given  price,  the  work  is  right. 

4.  A  man  mixed  four  kinds  of  oil,  worth  8s.,  9s.,  11s.,  and  12s. 
per  gal.  ;  the  mixture  was  worth  10s.  per  gal.  :  required,  the 
quantity  of  each. 


10 

° 


OBS.  1.  It  is  manifest  that  other  answers  maybe  obtained  by  connecting 
the  prices  in  a  different  manner. 

2.  It  is  also  manifest,  if  we  multiply  or  divide  the  answers  already  obtained 
oy  any  number,  the  results  will  fulfil  the  conditions  of  the  question  ;  conse- 
quently the  number  of  answers  is  unlimited. 

5.  A  goldsmith  has  gold  of  18,  20,  22,  and  24  carats  fine  :  how 
much  may  be  taken  of  each  to  form  a  mixture  21  carats  fine? 

CASE    Tl. 

555*  When  the  quantity  of  one  of  the  ingredients  and  the 
mean  price  of  the  mixture  are  given. 

Fir,d  the  difference  between  the  price  of  each  ingredient  and  the 
mean  price  of  tlie  required  mixture,  as  before  ;  then  by  proportion, 

As  the  difference  of  that  ingredient  whose  quantity  is  given,  is  to 
each  particular  difference,  so  is  the  quantity  given  to  tlie  quantity 
required  of  each  ingredient. 


ARTS.  555, 556.J  ALLIGATION.  359 

6.  How  many  pounds  of  sugar  at  10,  and  15  cents  a  pound, 
must  be  mixed  with  20  Ibs.  at  9  cents,  so  that  the  mixture  may 
be  worth  1 2  cents  a  pound  ? 

Solution. — Connecting  the  prices  as  directed,  the  differences 
between  them  and  the  mean,  are  3  cts.,  3  cts.  and  5  cts. 
Then  3  cts.  :  3  cts.  : :  20  Ibs.  :  to  the  Ibs.  at  9  cts. 
Also  3  cts.  :  5  cts.  : :  20  Ibs.  :       "       "       10  cts. 

Ans.  20  Ibs.  at  9  cts.,  and  33i  Ibs.  at  10  cts. 

7.  How  much  gold  of  16,  18,  and  22  carats  fine  must  be  mixed 
with  10  oz.  24  carats  fine,  that  the  mixture  maybe  20  carats  fine? 

8.  How  much  wool  at  20,  30,  and  24  cts.  a  pound  must  be  mixed 
with  95  Ibs.  at  50  cts.  to  form  a  mixture  worth  40  cts.  a  pound  ? 

CASE     III. 

556.  When  the  quantity  to  be  mixed  and  the  mean  price  of 
the  required  mixture  are  given. 

Find  the  difference  between  the  price  of  each  ingredient  and  the 
mean  price  of  the  required  mixture,  as  before  ;  then  by  proportion, 

As  the  sum  of  tJie  differences  is  to  each  particular  difference,  so 
is  the  whole  quantity  to  be  mixed,  to  the  quantity  required  of  each 
ingredient. 

9.  A  grocer  has  raisins  worth  8,  10,  and  16  cents  a  pound : 
how  many  of  each  kind  may  be  taken  to  form  a  mixture  of  112 
Ibs.  worth  12  cents  a  pound? 

Solution. — The  sum  of  the  differences  between  the  prices  of 

the  ingredients,  and  the  mean  price,  6  cts. +4  cts. +4  cts.=14  cts. 

Then  14  cts.  :  6  cts.  : :  112  Ibs.  :  to  the  Ibs.  at  16  cts. 

And    14  cts.  :  4  cts.  : :  112  Ibs.  :  to  the  Ibs.  at  8  and  10  cts. 

Ans.  48  Ibs.  at  16  cts.,  32  Ibs.  at  10  cts.,  and  32  Ibs.  at  8  cts. 

10.  How  much  wine  at  15,  17,  18,  and  22  shillings  per  gallon, 
may  be  mixed  to  form  a  mixture  of  320  gals,  worth  20  shillings 
per  gallon  ? 

11.  How  much  water  must  be  mixed  with  wine  worth  9s.  per 
gal.  to  fill  a  pipe,  so  that  the  mixture  may  be  worth  Vs.  per  gal.  ? 


360 


INVOLUTION. 


[SECT.  XVII, 


SECTION    XVII. 


INVOLUTION. 

Airr  557*  When  any  number  or  quantity  is  multiplied  into 
itself,  the  product  is  called  a  power.  Thus,  5X5  =  25;  3X3X3=- 
27;  2X2X 2X2  =  16;  the  products  25,  27,  and  16  are  powers. 

The  original  number,  that  is,  the  number  which  being  multi- 
plied into  itself,  produces  a  power,  is  called  the  root  of  all  the 
powers  of  that  number,  because  they  are  derived  from  it. 

558.  Powers  are  divided  into  different  orders ;  as  the  first, 
second,  third,  fourth,  fifth  power,  &c.  They  take  their  name  from 
the  number  of  times  the  given  number  is  used  as  a  factor,  in  pro- 
ducing the  given  power. 

OBS.  1.  The  first,  power  of  a  number  is  said  to  be  the  number  itself.  Strictly 
speaking,  it  is  not  a  power,  but  a  root.  (Art.  557.) 


3  yards. 


2.  The  second  power  of  a  number  is  also  called 
the  square;  (Art.  "257.  Obs.  1  ;)  for,  if  the  side  of  a 
square  is  3  yards,  then  the  product  of  3x3=9  yards,    . 
will  be  the  area  of  the  given  square.    (Art.  285.) -^ 
But  3X3=9  is  also  the  second  power  of  3 ;  hence,  it  ^ 
b  called  the  square.  M 

3.  The  diagonal  of  a  square  is  a  line  connecting 
two  of  its  opposite  corners. 


3.  The  third  power  of  a  number  is  also  called 
the  cube;  (Art.  258.  Obs.  1 ;)  for,  if  the  side  of  a 
cube  <s  3  feet,  then  the  product  of  3x3x3=27 
feex,  will  be  the  solidity  of  the  given  cube.  (Art. 
28G.)     But  3X3X3=27  is  also  the  'third  power 
of  3;  hence  it  is  called  the  cube.  Leg.  VII.  11.  Sen. 

4.  The  fourth  power  of  a  number  is  called  the 


3X3=9  yards. 
3  feet. 


I 


/ 


3X3X3=27  feet. 


QUEST.— 557.  What  is  a  power  ?  558.  How  are  powers  divided  ?  From  what  do  they 
take  their  name  7  Obs.  What  is  said  to  be  the  first  power  ?  What  is  the  second  powtl 
called  1  The  third?  The  fourth  1 


ARTS.  557-56 l.j  INVOLUTION.  3C1 

559.  Powers  are  denoted  by  a  small  figure  placed  above  the 
given  number  at  the  right  hand. 

This  figure  is  called  the  index  or  exponent.  It  shows  how  many 
times  the  given  number  is  employed  as  a  factor  to  produce  the 
required  power.  Thus, 

The  index  of  the  first  power  is  1 ;  but  this  is  commonly  omitted ; 
for,  (2)'==2. 

The  index  of  the  second  power  is  2  ; 

The  index  of  the  third  power  is  3  ; 

The  index  of  the  fifth  power  is  5  ;  &c.     That  is, 
21— 2,  the  first  power  of  2  ; 

22  =  2X2,  the  square,  or  second  power  of  2 ; 

23  =  2  X  2  X  2,  the  cube,  or  third  power  of  2 ; 

2<— 2X2X2X2,  the  biquadrate,  or  fourth  power  of  2; 

25=2  X  2X2X2X2,  the  fifth  power  of  2  ; 

21 =2X2X2X2X2X2,  the  sixth  power  of  2  ;  &c. 

Ex.  1.  Express  the  square  of  17,  and  the  cube  of  19. 

Ans.  17',  19. 

Express  the  given  powers  of  the  following  numbers : 

2.  The  square  of  54.  7.  The  2d  power  of  299. 

3.  The  cube  of  43.  8.  The  4th  power  of  785. 

4.  The  square  of  87.  9.  The  5th  power  of  228. 
K.  The  biquadrate  of  91.             10.  The  8th  power  of  693. 
6.  The  3d  power  of  41G.             11.  The  32d  power  of  999. 

5 GO.  The  process  of  finding  a  power  of  a  given  number  by 
multiplying  it  into  itself,  is  called  INVOLUTION. 

5G1.  Hence,  to  involve  a  number  to  any  required  power. 

Multiply  the  given  number  into  itself,  till  it  is  taken  as  a  factor, 
as  many  times  as  there  are  units  in  the  index  of  the  power  to  which 
the  number  is  to  be  raised.  (Art.  558.) 

OBS.  1.  The  number  of  multiplications  in  raising  a  number  to  any  rriveri 
power,  is  one  less  than  the  imlex  of  the  required  power.     Thus,  32—3x3;  th 
3  is  taken  twice  as  a  factor,  but  there  is  but  one  multiplication. 

UCKST. — .V>".  How  are  powers  denoted  ?  What  is  this  figure  called  1  What  does  it 
show  ?  What  is  the  index  of  the  first  power!  Of  the  second?  The  third?  Fifth  1 
SoO.  What  is  involution  7  561.  How  is  a  number  involved  to  any  required  power  1 


362  INVOLUTION.  [SECT.  XVD 

2.  A  Fraction  is  raised  to  a  power  by  multiplying  it  into  itself.     Thus,  th« 
square  of  |  is  -f  xl=f« 

Mixed  numbers  should  be  reduced  to  improper  fractions,  or  the  common 
fraction  to  a  decimal.  They  may  however  be  involved  without  reducing  them. 
(Art.  2-20.  Obs.) 

3.  The  process  of  raising  a  number  to  a  high  power,  may  often  be  contracted 
by  multiplying  together  powers  already  found.     The  index  of  the  power  thus 
found,  is  equal  to  the  sum,  of  the  indices  of  the  powers  multiplied  together. 
Thus,  2X2=4;  and  4x4=2x2X2x2,  or  2*. 

i  35;  and 


12.  What  is  the  square  of  23  ? 

Common  Operation.  Analytic  Operation. 

23  23  =  2  tens  or  20+3  units. 

23  23  =  2  tens  or  20  +  3  units. 
69  60+9 

46^  400+   60  __ 

"629  A  *s.  And  400  +  120+9  =  529.  Ans. 

It  will  be  seen  from  this  operation  that  the  square  of  20  +  3 
contains  the  square  of  the  first  part,  viz  :  20X20  =  400,  added  to 
twice  the  product  of  the  two  parts,  viz:  20X3  +  20X3  =  120, 
added  to  the  square  of  the  last  part,  viz  :  3X3  =  9.  Hence, 

562.  The  square  of  the  sum  of  two  numbers  is  equal  to  the 
square  of  the  first  part,  added  to  twice  the  product  of  the  two 
parts,  and  the  square  of  the  last  part. 

OBS.  1.  The  product  of  any  two  factors  cannot  have  more  figures  than  both 
factors,  nor  but  one  less  than  both.  For  exampfe,  take  9,  the  greatest  num- 
ber which  can  be  expressed  by  one  figure.  (Art.  34.)  And  (9)2,  or  9X9=81, 
has  two  figures,  the  same  number  which  both  factors  have.  99  is  the  greatest 
number  which  can  be  expressed  by  two  figures  ;  (Art,  34  ;)  and  (99)2,  or  99X 
99=9eOl,  has  four  figures,  the  same  as  both  factors  have. 

Again,  1  is  the  smallest  number  expressed  by  one  figure,  and  (J)2,  or  1X1 
=  1,  has  but  one  figure  less  than  both  factors.  10  is  the  smallest  number 
which  can  be  expressed  by  two  figures;  and  (10)2,  or  10X10=100,  has  one 
figure  less  than  both  factors.  Hence, 

QUEST.  —  Obs.  How  many  multiplications  are  there  in  raising  a  number  to  a  given 
power  1  How  is  a  fraction  involved  ?  A  mixed  number  7  562.  What  is  the  square  o/ 
the  sum  of  two  numbers  equal  to?  Obs.  How  many  figures  are  there  in  the  product  of 
any  two  factors  1  How  many  figures  will  the  square  of  a  number  contain  ?  The  cube  ? 


ARTS.  562-564'.]  EVOLUTION.  363 

2.  A  sqv^are  cannot  have  more  figures  than  double  the  number  of  the  root  of 
first  power,  nor  bid  one  less. 

3.  A  cube  cannot  have  more  figures  than  triple  the  number  of  the  root  or  Jivst 
pnver,  nor  but  two  less. 

4.  All  powers  of  1  are  the  same,  viz:  1 ;  for,  1X1X1X1,  &c.=l. 

13.  What  is  the  square  or  second  power  of  123  ? 

14  The  cube  of  135  ?  23.  The  cube  of  .012  ? 

15.  The  square  of  2880  ?  24.  The  square  of  .00125  ? 

16.  The  4th  power  of  10  ?  25.  The  square  of  -f  ? 

17.  The  5th  power  of  5  ?  26.  The  cube  of  i? 

18.  The  7th  power  of  6  ?  27.  The  square  of  •££? 

19.  The  6th  power  of  7  ?  28.  The  cube  of  fW? 

20.  The  8th  power  of  4  ?  29.  The  square  of  4£  ? 

21.  The  9th  power  of  9?  30.  The  square  of  7|? 

22.  The  souare  of  2.5  ?  31.  The  square  of  38-JI? 

EVOLUTION. 

503.  If  we  resolve  25  into  two  equal  factors,  viz:  5  and  5, 
each  of  these  equal  factors  is  called  a  root  of  25.  So  if  we  re- 
solve 27  into  three  equal  factors,  viz :  3,  3,  and  3,  each  factor  is 
called  a  root  of  27  ;  if  we  resolve  16  into  four  equal  factors,  viz : 
2,  2,  2,  and  2,  each  factor  is  called  a  root  of  16.  And,  universally, 
when  a  number  is  resolved  into  any  number  of  equal  factors,  each 
of  those  factors  is  said  to  be  a  root  of  that  number.  Hence, 

564.  A  root  of  a  number  is  a  factor,  which,  being  multiplied 
into  itself  a  certain  number  of  times,  will  produce  that  number. 

OBS.  Roots,  as  well  as  powers,  are  divided  into  different  orders.  Thus, 
when  a  number  is  resolved  >nto  two  equal  factors,  each  of  these  factors  is 
called  the  second  or  square  root ;  when  resolved  into  three  equal  factors,  each 
of  these  factors  is  called  the  third  or  cube  root,  &c.  Hence, 

The  name  of  the  root  expresses  the  number  of  ejiual  factors  into  which  the  given 
number  is  to  be  resolved. 


Roots.       | 

1|  2 

1    3|    4| 

5  '      6 

25]    36 

1      7 

8  |      9 

1    10 

1   11 

12 

Squares.  ' 

1  1  4 

1    9  1  16  | 

|    49 

64  |    81 

|    100 

|    121 

144 
~i  1728 

Cubes.      | 

1|  8 

|  27  |  64  ' 

125 

|  216 

|  343 

512  |  729 

|  1000 

|  1331 

QUEST.— Ofo.  What  are  all  powers  of  J 1   564.  What  is  a  root  if  a  number  1    Obj.  Wha/ 
does  the  name  of  the  root  express  ? 

16* 


364  EVOLUTION.  [SECT.  XVII 

565*  The  process  of  resolving  numbers  into  equal  factors  is 
called  EVOLUTION,  or  tlie  Extraction  of  Roots. 

OBS.  1.  Evolution  is  the  opposite  of  involution.  (Art.  500.)  One  is  finding 
a  power  of  a  number  by  multiplying  it  into  itself;  the  other  is  finding  a  root  by 
resolving  a  number  into  equal  factors.  Powers  and  roots  are  therefore  correla- 
tive terms.  If  one  number  is  a  power  of  another,  the  latter  is  a  root  of  the  for- 
mer. Thus,  27  is  the  cube  of  3  ;  and  3  is  the  cube  root  of  27. 

2.  The  learner  will  be  careful  to  observe,  that 

In  subtraction,  a. number  is  resolved  into  two  parts; 

In  division,  a  number  is  resolved  into  two  factors; 

In  evolution,  a  number  is  resolved  into  equal  factors. 

566*  Roots  are  expressed  in  two  ways  ;  one  by  the  radical 
sign  (y)  placed  before  a  number ;  the  other  by  &  fractional  index 

placed  above  the  number  on  the  right  hand.  Thus,  ^/4,  or  42 
denotes  the  square  or  2d  root  of  4  ;  V27,  or  27 3  denotes  the  cube 
.or  3d  root  of  27 ;  VI 6,  or  164  denotes  the  4th  root  of  16. 

OBS.  1.  The  figure  placed  over  the  radical  sign,  denotes  the  root,  or  the  num- 
ber of  equal  factors  into  which  the  given  number  is  to  be  resolved.  The  figure 
for  the  square  root  is  usually  omitted,  and  simply  the  radical  sign  -^  is  placed 
before  the  given  number.  Thus  the  square  root  of  25  is  written  ^/~25. 

2.  When  a  root  is  expressed  by  a  fractional  index,  the  denominator,  like  the 
figure  over  the  radical  sign,  denotes  the  root  of  the  given  number.     Thus, 
;25)*  denotes  the  square  root  of  25 ;  (27)*  denotes  the  cube  root  of  27. 

3.  A  fractional  index  whose  numerator  is  greater  than  1,  is  sometimes  used. 
In  such  cases  the  denominator  denotes  the  root,  and  the  numerator  the  power 

of  the  given  number.  Thus,  8'  denotes  the  square  of  the  cube  root  of  8,  or 
the  cube  root  of  the  square  of  8,  each  of  which  is  4. 

4.  The  radical  sign  ^ ,  is  derived  from  the  letter  r,  the  initial  of  the  Latin 
radix,  a  root. 

1.  Express  the  cube  root  of  74.     Ans.  \/74,  or  74 3. 

2.  The  square  root  of  119.  5.  The  square  root  of  £. 

3.  The  4th  root  of  231*  6.  The  cube  root  of  f. 

4.  The  9th  root  of  685.  7.  The  4th  root  of  \\. 

6,  Express  the  3d  power  of  th*  4th  root  of  6.     Ans.  6*. 
9.  Express  the  2d  power  of  the  3d  root  of  81. 

QUEST.— 565.  What  is  evolution?  Obs.  Of  what  is  it  the  opposite?  Into  what  are 
nambers  resolved  in  subtraction?  In  division?  In  evolution?  566.  How  many  ways 
are  roots  expressed  ?  What  are  they  ?  Obs.  What  does  the  figure  over  the  radical  sign 
denote  ?  What  the  denominator  of  the  fractional  index  ? 


ARTS.  565-570.]       SQUARES  AND  CUBES.  365 

5G7.  A  number  which  can  be  resolved  into  equal  factors,  or 
whose  root  can  be  exactly  extracted,  is  called  a  perfect  power,  and 
its  root  is  called  a  rational  number.  Thus,  1C,  25,  27,  &c.,  are 
perfect  powers,  and  their  roots  4,  5,  3,  are  rational  numbers. 

568.  A  number,  which  cannot  be  resolved  into  equal  factors, 
or  whose  root  cannot  be  exactly  extracted,  is  called  an  imperfect 
power  ;  and  its  root  is  called  a  Surd,  or  irrational  number.    Thus, 
15,  17,  45,  <fec.,  are  imperfect  powers,  and  their  roots  3.8  + ',  4.1  +; 
6.7+,  &c.,  are  surds,  for  their  roots  cannot  be  exactly  extracted. 

OBS.  A  number  may  be  a  perfect  power  of  one  degree  and  an  imperfect 
power  of  another  degree.  Thus,  10  is  a  perfect  power  of  the  second  degree, 
but  an  imperfect  power  of  the  third  degree  ;  that  is,  it  is  a  perfect  square  but 
not  a  perfect  cube.  Indeed  numbers  are  seldom  perfect  powers  of  more  than 
one  degree.  1G  is  a  perfect  power  of  the  2d  and  4th  degrees;  64  is  a  perfect 
power  of  the  2d,  3d  and  Gth  degrees. 

569.  Every  rooi,  as  well  as  power  of  l,is  1.  (Art.  5G2.  Obs.  4.) 
Thus,  (I)2,  (I)3,  (I)6,  and  -/I,  ^1,  V%  &c.,  are  all  equal. 

PROPERTIES    OF   SQUARES   AND   CUBES. 

5  TO.  The  properties  of  numbers  in  general,  have  already  been 
given.  The  following  pertain  to  square  and  cubic  numbers. 

1.  The  product  of  any  two  or  more  square  numbers,  is  a  square;  and  the 

2  -2 

product  of  any  two  or  more  cubic  numbers,  is  a  cube.     Thus  2   X3   =3G,  the 
square  of  G;  and  2   X3   =21G,  is  the  cube  of  6. 

2.  If  a  square  number  is  divided  by  a  square,  the  quotient  will  be  a  square. 
Thus,  144-5-9  =  Hi,  which  is  the  square  of  4. 

3.  If  a  square  number  is  either  multiplied  or  divided  by  a  number  that  is  not 
a  square,  neither  the  product  nor  quotient  will  be  a  square. 

4.  If  you  double  the  number  of  times  a  number  is  taken  as  a  factor,  it  will 
not  produce  double  the  product,  but  the  square  of  it.     Thus,  3x3^9,  and  3x3 
X3,<3=81,  and  not  18. 

5.  The  product  of  two  different  prime  numbers  cannot  be  a  square. 

G.  The  product  of  no  two  different  numbers,  which  are  prime  to  each  otler, 
will  make  a  square,  unless  each  of  those  numbers  is  a  square. 

7.  The  square  and  cube  of  an  even  number  are  eren. ;  and  the  square  am. 
cube  of  an  odd  number  are  odd.  (Art.  IGl.  Prop.  G,  10.)  Hence, 

QI-KST. — 507.  What  is  a  perfect  power?    What  is  a  rational  number  1    508.  An  imper- 
.  feet  jM>wer  7   A  surd  1    Obs.  Are  numbers  ever  perfect  powers  of  one  degree  und  imperfee 
powers  of  another  degree  7    5(59.  What  are  all  roots  and  powers  of  1 1 


366  SQUARES    AND    CUBES.  [SECT.  XV11, 

8.  The  square  or  cube  root  of  an  even  number,  is  even ;  and  the  square  or 
^ubc  root  of  an  odd  number,  is  odd. 

9.  Every  square  number  necessarily  ends  with  one  of  these  figures,  1,  4,  5, 
69;  or  with  an  croi  number  of  ciphers  preceded  by  one  of  these  figures. 

10.  No  number  is  a  square  that  ends  in  2,  3,  7,  or  8. 

11.  A  cubic  number  may  end  in  any  of  the  natural  numbers,  1,  2,  3,  4,  5,  6 
7,  8,  9,  or  0. 

12.  All  the  powers  of  any  number,  ending  in  5.  will  also  end  in  5;  and  if 
a  number  ends  in  6,  all  its  powers  will  end  in  6. 

13.  E/;ry  square  number  is  divisible  by  3,  and  also  by  4,  or  becomes  so 
when  diminished  by  unity.     Thus,  4,  9,  ib',  25,  &c.,  are  all  divisible  by  3,  and 
by  4,  or  become  so  when  diminished  by  1. 

14.  Every  square  number  is  divisible  also  by  5,  or  becomes  so  when  increased 
o  •  diminished  by  unity.     Thus,  3b' — 1,  and  49-f  I,  are  divisible  by  5. 

15.  Any  even  square  number  is  divisible  by  4. 

16.  An  odd  square  number,  divided  by  4,  leaves  a  remainder  of  1. 

17.  Every  odd  square  number,  decreased  by  unity,  is  divisible  by  8. 

18.  Evciy  number  is  either  a  square,  or  is  divisible  into  two,  or  three,  or  Jon* 
squares.     Thus  30  is  equal  to  25-f-4+l ;  33=16-j-lO-|-l ;  63=49-f-9-j-4-j-l. 

19.  The  product  of  the  sum  and  difference  of  two  numbers,  is  equal  to  the 
difference  of  their  squares.     Thus,  (5+3)  X  (5— 3)  =  16 ;  also  5«— 3*=  10. 

20.  If  two  numbers  are  such,  that  their  squares,  when  added  together,  form 
a  square,  the  product  of  these  two  numbers  is  divisible  by  6.     Thus,  3  and  4, 
the  sum  of  whose  squares,  9-j-16=25,  is  a  square  number,  and  their  product 
12,  is  divisible  by  6.     Hence, 

21.  To  find  two  numbers,  the  sum  of  whose  squares  shall  be  a  square  number. 

Take  any  two  numbers  and  multiply  Uu:m*togclhcr ;  Hue  double  of  their  prod- 
uct will  be  one  of  the  numbers  sought,  and  the  difference  of  their  squares  will  be 
the  other.  Thus,  take  any  two  numbers,  as  2  and  3 ;  the  double  of  their  prod- 
uct is  12,  and  he  difference  of  their  squares  is  5;  now  122-J-52=169,  the 
square  of  13. 

22.  When  two  numbers  are  such,  that  the  difference  of  their  squares  is  a 
square  number ;  the  sum  and  difference  of  these  numbers  are  themselves  square 
numbers,  or  the  double  of  square  numbers.     Thus,  8  and  10  give  for  the  dif- 
ference of  their  squares  30  ;  and  18,  the  sum  of  these  numbers,  is  the  double 
of  9,  which  is  a  square  number,  and  2,  their  difference,  is  the  double  of  1,  which 
is  also  a  square  number. 

23.  If  two  numbers,  the  difference  of  which  is  2,  be  multiplied  together,  their 
product  increased  by  unity,  will  be  the  square  of  the  intermediate  number. 

21.  The  sum  01  difference  of  two  numbers,  will  measure  the  difference  of 
heir  squares. 

25.  The  sum  of  two  numbers,  differing  by  unity,  is  equal  to  the  difference 
of  their  squares. 

26.  The  sum  of  two  numbers  will  measure  the  sum  of  their  cubes;  and  th« 
difference  of  two  numbers  will  measure  the  difference  of  their  cubes. 


ART.  571.]  SUUARE  ROOT.  3£7 

27.  If  a  square  measures  a  square,  or  a  cube  a  cube,  the  root  will  al"o  -.neas* 
urc  the  root. 

28.  If  one  number  is  prime  to  another,  its  square,  cube,  &c.,  will  also  be 
prime  to  it. 

29.  The  difference  between  an  integral  cube  and  its  root,  is  always  divisi 
ble  by  6. 

30.  If  any  series  of  numbers  beginning  from  I,  be  in  continued  geometrical 
proportion,  the  3d,  5th,  7th,  &c.,  will  be  squares;    the  4th,  7th,  10th,  &c., 
cubes  ;    and  the  7th  will  be  both  a  square  and  a  cube.     Thus,  in  the  series, 
1,  2,  4,  8,  16,  32,  G4,  &c.,  the  3d,  5th,  and  7th  terms  are  squares;  the  4th  am/ 
7th  are  cubes ;  and  the  7th  is  both  a  square  and  a  cube. 


EXTRACTION  OF  THE  SQUARE  ROOT. 

571.  To  extract  tJie  SQUARE  ROOT,  is  to  resolve  a  given  number 
into  two  equal  factors  ;  or,  to  find  a  number  which  being  multiplied 
into  itself,  will  produce  the  given  number.  (Art.  564.  Obs.) 

Ex.  1.  What  is  the  square  root  of  36  ? 

Solution. — Resolving  the  given  number  into  two  equal  factors, 
we  have  36  =  6X6.  Ans.  The  square  root  of  36  is  6. 

2.  What  is  the  length  of  one  side  of  a  square  field  whi^h  con- 
tains 529  square  rods? 

Operation.  Since  we  may  not  see  what  the  root  of  529 

529(23         is  at  once,  we  will  separate  it  into  two  periods 
4  by  placing  a  point  over  the  9  and  another  over 

43)129  the  5.     Now  the  greatest  square  of  5,  the  left 

129  hand  period,  is  4,  the  root  of  which  is  2.     Plac- 

ing the  2  on  the  right  of  the  number,  we  sub- 
tract its  square  from  the  period  5,  and  to  the  right  of  the  re- 
mainder bring  down  the  next  period.  We  then  double  the  2,  the 
part  of  the  root  already  found,  and,  placing  it  on  the  left  of  the 
dividend  for  a  partial  divisor,  we  perceive  it  is  contained  in  the 
dividend,  omitting  its  right  hand  figure,  3  times.  Placing  the  3 
on  the  right  of  the  root,  also  on  the  right  of  the  partial  divisor, 
we  multiply  the  divisor  thus  completed  by  3,  and  subtract  the 
product  from  the  dividend.  The  answer  is  23  rods/ 

QUEST.— 571.  What  is  it  to  extract  the  square  root  of  a  number  7 


368  saUARE  ROOT.  [SECT.  XVII 

bate. — Since  the  root  is  to  contain  2  figures,  the  2  stands  in  tens  place, 
hence  the  first  part  of  the  root  found  is  properly  20 ;  which  being  doubled, 
gives  40  for  the  divisor.  For  convenience  we  omit  the  cipher  on  the  right ; 
and  to  compensate  for  this,  we  omit  the  right  hand  figure  of  the  dividend. 
This  is  the  same  as  dividing  both  the  divisor  and  dividend  by  10,  and  therefore 
does  not  alter  the  quotient.  (Art.  146.) 

572.  Hence,  we  derive  the  following  general 

RULE  FOR  EXTRACTING  THE  SQUARE  ROOT. 

I.  Separate  the  given  number  into  periods  of  two  figures  each,  by 
placing  a  point  over  the  units  figures,  then  over  every  second  fig- 
ure towards  the  left  in  whole  numbers,  and  over  every  second  figure 
towards  the  right  in  decimals. 

II.  Find  the  greatest  square  number  in  the  first  or  left  hand 
period,  and  place  its  root  on  the  right  of  the  number  for  the  first 
figure  in  the  root.     Subtract  the  square  of  this  figure  of  the  root 
from  the  period  under  consideration  ;  and  to  the  right  of  the  re- 
mainder bring  down  the  next  period  for  a  dividend. 

III.  Double  the  root  just  found  and  place  it  on  the  left  of  the 
dividend  for  a  partial  divisor  ;  find  how  many  times  it  is  contained 
in. the  dividend,  omitting  its  right  hand  figure ;  place  the  quotient 
on  the  right  of  the  root,  also  on  the  right  of  the  partial  divisor ; 
multiply  the  divisor  thus  completed  by  the  figure  last  placed  in  the 
root ;  subtract  the  product  from  the  dividend,  and  to  the  remainder 
bring  down  the  next  period  for  a  new  dividend. 

IV.  Double  the  root  already  found  for  a  new  partial  divisor,  di- 
vide, &c.,  as  before,  and  thus  continue  tlie  operation  till  the  root  of 
all  the  periods  is  extracted. 

If  there  is  a  remainder  after  all  the  periods  are  brought  down, 
the  operation  may  be  continued  by  annexing  periods  of  ciphers. 

PROOF. — Multiply  the  root  into  itself ;  and  if  the  product  it 
equal  to  the  given  number,  the  work  is  right.  (Art.  5G4.) 

573  Demonstration. — Take  any  number  as  that  in  the  last  example ;  thf  n 
separating  it  into  parts,  529=500-|-29.  Now  the  greatest  square  in  500  is  -100 
the  root  of  which  is  20,  with  a  remainder  of  100 ;  consequently,  the  first  part  ol 

QI-KST.— 572.  What  is  the  first  step  in  extr.-ic  )g  the  sqtnre  root  ?  The  second  ?  Thinlt 
Fourth  3  When  there  is  a  remainder,  how  proceed  1  How  is  the  square  root  proved  1 


ARTS.  572-574.  [  SUI/ARE  ROOT.  309 

the  root  must  be  20,  and  the  true  remainder  is  lOO-f-29,  or  129.  And  since 
there  are  three  figures  in  the  given  number,  there  must  be  two  figures  in  the 
root;  (.Art.  562.  Obs. 2;)  but  the  square  of  the  sum  of  two  numbers,  is  equal 
to  the  square  of  the  first  part  ad  'ed  to  twice  the  product  of  the  two  parts  and 
the  square  of  the  last  part;  it  follows  therefore  that  the  remainder  129,  must 
be  t  vice  the  product  of  20  into  the  part  of  the  root  still  to  be  found,  together 
with  the  square  of  that  part.  (Art.  562.)  Now  dividing  129  by  40  the  double 
of  20,  the  quotient  is  3,  which  being  added  to  40  makes  43 ;  finally,  multiply- 
ing 43  by  3,  the  product  is  129,  which  is  manifestly  twice  the  product  of  20 
into  3,  together  with  the  square  of  3.  In  the  same  manner  the  operation  may 
be  proved  in  every  case.  (For  illustration  of  this  rule  by  geometrical  figures, 
*ce  Practical  Arithmetic,  £.  318.) 

1.  The  reason  for  separating  the  given  numbers  into  periods  of  two  figures 
each,  is  that  a  square  number  can  not  have  more  figures  than  double  the  num- 
ber oi  figures  in  the  root,  nor  but  one  less.     It  also  shows  how  many  figures  the 
root  will  contain,  and  thus  enables  us  to  find  part  of  it  at  a  time.  (Art.  562. 
Obs.  2.) 

2.  The  reason  for  doubling  that  part  of  the  root  already  found  for  a  divisor, 
is  because  the  remainder  is  double  the  product  of  the  first  part  of  the  root  into 
the  second  part,  together  with  the  square  of  the  second  part. 

3.  In  dividing,  the  right  hand  figure  of  the  dividend  is  omitted,  because  the 
cipher  on  the  right  of  the  divisor  being  omitted,  the  quotient  would  be  10 
times  too  large  for  the  next  figure  in  the  root.  (Arts.  130,  146.) 

4.  The  last  figure  of  the  root  is  placed  on  the  right  of  the  divisor  simply  fot 
convenience  in  multiplying  it  into  itself. 

OBS.  1.  The  product  of  the  divisor  completed  into  the  figure  last  placed  in 
the  root,  cannot  exceed  the  dividend.  Hence,  in  finding  the  figure  to  be  placed 
in  the  root,  some  allowance  must  be  made  for  carrying,  when  the  product  of 
this  figure  into  itself  exceeds  9. 

2.  If  the  right  hand  period  of  decimals  is  deficient,  it  must  be  completed  by 
annexing  a  cipher  to  it. 

3.  There  will  always  be  as  many  decimal  figures  in  the  root,  as  there  are 
periods  of  decimals  in  the  given  number. 

574*  The  square  root  of  a  common  fraction  is  found  by  ex- 
tracting the  root  of  the  numerator  and  denominator. 

A  mixed  number  should  be  reduced  to  an  improper  fraction. 
When  either  the  numerator  or  denominator  of  a  common  fraction 
is  not  a  perfect  square,  the  fraction  may  be  reduced  to  a  decimal, 
and  the  approximate  root  be  found  as  above. 

QUEST.— 573.  Dem.  Why  do  we  separate  the  given  number  into  periods  of  two  figures 
each  1  Why  double  the  root  thus  found  for  a  divtaor ?  Why  omit  the  right  hand  figure 
of  the  dividend  ?  Why  place  the  last  figure  of  the  root  on  the  right  of  the  divisor?  Obs 
How  many  decimal  figures  will  there  be  in  the  root?  574.  How  is  the  square  roc*  of  a 
common  fraction  found  ?  Of  a  mixed  number  ? 


370 


SQUARE    ROOT. 


XVII. 


"Required  the  square  root  of  the  following  numbers  : 


3. 

2601. 

10. 

27889 

17. 

566.44. 

24. 

If. 

4. 

5329. 

11. 

961. 

18. 

7.3441. 

25. 

iff 

5. 

784. 

12. 

97. 

19. 

.81796. 

26. 

f. 

6, 

87. 

13. 

7. 

20. 

1169.64. 

27. 

m. 

7 

4761. 

14. 

190. 

21. 

627264. 

28. 

794£. 

8. 

7056. 

15. 

43681. 

22. 

3.172181. 

29. 

2071* 

9. 

.9801. 

16. 

47089. 

23. 

10342656. 

30. 

34967  jV. 

31.  What  is  the  square  root  of  152399025  ? 

32.  What  is  the  square  root  of  119550669121  ? 

33.  What  is  the  square  root  of  964.5192360241  ? 

575.  When  the  root  is  to  be  extracted  to  many  figures,  tha 
operation  may  be  contracted  in  the  following  manner.1 

First  find  half,  or  one  more  than  half  the  number  of  figures  re' 
quired  in  the  root  ;  then  having  found  tlie  next  true  divisor,  cut  off 
its  right  hand  figure,  and  divide  the  remainder  by  it  ;  place  Hi* 
quotient  in  the  root,  and  continue  the  operation  as  in  contraction  of 
division  of  decimals.  (Art.  333.) 

34.  Required  the  square  root  of  365  to  eleven  figures  in  the 
root.  Ans.   19.104973174. 

35.  Required  the  square  root  of  2  to  twelve  figures. 

36.  Required  the  square  root  of  3  to  seventeen  figures. 

APPLICATIONS   OF   THE    SQUARE   ROOT. 

576.  A  triangle  is  a  figure  which  has  three  sides  and  three 
angles.     When  one  of  the  sides  of  a  triangle  is  perpendicular  to 
another  side,  the  angle  between  them  is  called  a  right-angle. 

C 

577.  A  right-angled    triangle  is  a 
triangle  which  has  a  right-angle. 

The  side  opposite  the  right-angle  is 
called  the  hypothenuse,  and  the  other 
two  sides,  the  base  and  perpendicular. 
The  triangle  ABC  is  right-angled  at  B, 
and  the  side  AC  is  the  hypothenuse. 


ART?.  575-580.J  SQUARE  ROOT 


371 


578.  The  square  described  on  the  hypothenuse  of  a  right- 
angled  triangle,  is  equal  to  the  sum  of  the  squares  described  on 
the  other  two  sides.  (Thomsons  Legendre,  B.  IV.  11,  Euc.  I.  47.) 

The  truth  of  this  principle  may  be  seen  from  the  following  geometrical  illus- 
tration. Thus, 

Let  the  base  AB  of  the  right- 
angled  triangle  ABC  be  4  feet, 
he  perpendicular  AC,  be  3  feet  ; 

hen  will  the  squares  described 

n  the  base  AB,  and  the  per- 
pendicular AC,  contain  as  many 
square  feet  as  the  square  de- 
scribed on  the  hypothenuse  BC. 
Now  (4)«-|-(3)a=25  sq.  ft.  ;  and 
the  square  described  on  BC  also 
contains  25  sq.  ft.  Hence,  the 
square  described  on  the  hypothe- 
nuse of  any  right-angled  trian- 
gle, is  equal  to  the  sum  of  the 
squares  described  on  the  other  two 
sides. 

OBS.  Since  the  square  of  the  hypothenuse  BC,  is  25,  it  follows  that  the 
or  5,  must  be  the  hypothenuse  itself.     Hence, 


579,  When  the  base  and  perpendicular  are  given,  to  find  the 
hypothenuse. 

Add  the  square  of  the  base  to  the  square  of  ike  perpendicular  ', 
the  square  root  of  the  sum  will  be  the  hypothenuse. 

Thus,  in  the  Yight-angled  triangle  ABC,  if  the  base  is  4  and 
the  perpendicular  3,  then  (4)2+(3)2=25,  and  \/25  =  5,  the  hypo- 
thenuse. 

580.  When  the  hypothenuse  and  base  are  given,  to  find  the 
perpendicular. 

From  the  square  of  the  hypothenuse  subtract  the  square  of  the 
base,  and  the  square  root  of  the  remainder  will  be  the  perpendicular. 

QUKST.—  576.  What  is  a  triangle  ?  What  is  a  right-angle  ?  577.  What  is  a  right 
angled  triangle  ?  What  is  the  side  opposite  the  right-angle  called  7  What  are  the  other 
two  sides  called  ?  578.  What  is  the  square  described  on  the  hypothenusf  «-ual  tc1 

579.  When   the  base  and   perpendicular  are  given,  how  is  the   hypothenuse  found  \ 

580.  Whei  the  hypothenuse  and  base  are  given,  how  is  the  perpendicular  found  7 


372  sauARE  ROOT.  [SECT.  XV11. 

Thus,  if  the  Iwpothenuse  is  5  and  the  base  4,  then  (o)2  —  (4)* 
=  9,  and  y/Q  =  S,  the  perpendicular. 

581.  When  the  hypothenuse  and  the  perpendicular  are  given, 
to  find  the  base. 

from  the  square  of  the  hypothenuse  subtract  the  square  of  the  per- 
pendicular, and  the  square  root  of  the  remainder  will  be  the  base. 

Tl  us,  if  the  hypothenuse  is  5  and  the  perpendicular  3,  then 
(o)2  —  (3)2  =  16,  and  Vl6=4,  the  base. 


OBS.  1.  From  the  preceding  principles  it  is  manifest  that  the  area  >f  a  square 
may  be  found  by  dividing  the  square  of  its  hypothenuse  by  2.  (Arts.  '285,  578.) 

2.  The  areas  of  all  similar  figures  are  to  each  other  as  tHe  squares  of  their 
homologous  sides,  or  their  like  dimensions.  (Leg.  IV.  25,  27.  V.  10.)     Hence, 

The  s:  .n  of  the  areas  of  equilateral  or  other  similar  triangles,  also  of  similar 
polygons,  circles  and  semicircles  described  on  the  base  and  perpendicular  of  a 
right-angled  triangle,  is  equal  to  the  area  of  a  similar  figure  described  on  the 
hypothenuse. 

3.  The  square  of  a  simple  ratio  is  called  a  duplicate  ratio  ;  the  cube  of  a  sim- 
ple ratio,  a  triplicate  ratio.  * 

The  ratio  of  the  square  roots  of  two  numbers  is  called  a  sub-duplicate  ratio; 
that  of  the  cube  roots,  a  sub-triplicate  ratio. 

Ex.  1.  If  a  street  is  28  feet  wide,  and  the  height  of  a  tower  is 
96  feet,  how  long  must  a  rope  be  to  reach  from  the  top  of  the 
to  wer  to  the  opposite  side  of  the  street  ? 

Solution.—  (96)2  +  (28)2=:10000,  and  VlOOOO=100  ft. 

2.  A  ladder  40  feet  long  being  placed  at  the  opposite  side  of  a 
street  24  feet  wide,  just  reached  the  top  of  a  house  :  how  high 
was  the  house  ? 

3.  Two  ships,  one  sailing  7  miles,  the  other  12  miles  an  hour, 
spoke  each  other  at  sea  ;  one  was  going  due  east,  the  other  due 
south  :  how  far  apart  were  they  at  the  expiration  of  12  hours  ? 

4.  What  is  the  length  of  the  side  of  a  square  farm  which  con- 
tains 300  acres;  and  how  far  apart  are  its  opposite  corners? 

582.  A  mean  proportional  between  two  numbers  is  equal  io 
the  square  root  of  their  product.  (Arts.  494,  498.  Obs.  2.) 

QrtsT.—  .W  When  the  hypothenuse  and  perpendicular  are  given,  how  is  the  base 
found? 


ARTS.  581-584.]  SCIUARE  ROOT  373 

5.  Find  a  mean  proportional  between  2  ind  8. 
Solution. — 8X2  =  16;  and  >/16=4.  Ans. 

Find  a  mean  proportional  between  the  following  numbers: 

6.  4  and  25.  10.   121  and  36.  14.  \  and  if. 

7.  9  and  36.  11.   196  and  144.  15.  |f  and  -£fa. 

8.  16  and  81.  12.  2.56  and  49.  16.  H  and  iff. 

9.  64  and  25.  13.  6.25  and  729.         17,  -ff  and  iff-. 

583*  To  find  the  side  of  a  square  equal  in  area  to  any  given 
surface. 

Extract  the  square  root  of  the  given  area,  and  it  will  be  the  side 
of  the  square  sought. 

OBS.  When  it  is  required  to  find  the  dimensions  of  a  rectangular  field,  equal 
in  area  to  a  given  surface,  and  whose  length  is  double,  triple,  or  quadruple, 
&c.,  of  its  breadth,  the  square  root  of  £,  £,  {,  of  the  given  surface,  will  be  the 
width ;  and  this  being  doubled,  tripled,  or  quadrupled,  as  the  case  may  be, 
will  be  the  length. 

18.  What  is  the  side  of  a  square  equal  in  area  to  a  rectangular 
field  81  rods  long,  and  49  rods  wide? 

19.  What  is  the^side  of  a  square  equal  in  area  to  a  triangular 
field  which  contains  160  acres? 

20.  What  is  the  side  of  a  square  equal  in  area  to  a  circular 
€eld  which  contains  640  acres  ? 

21.  What  are  the  length  and  'breadth  of  a  rectangular  field 
which  contains  480  acres,  and  whose  length  is  triple  its  breadth  ? 

22.  A  general  arranged  10952  soldiers,'so  that  the  number  in 
rank  was  double  the  file :  how  many  were  there  in  each  ? 

584.  When  the  sum  of  two  numbers  and  the  difference  of 
their  squares  are  given,  to  find  the  numbers. 

Divide  the  difference  of  their  squares  by  the  sum  of  the  numbers^ 
and  the  quotient  will  be  their  difference;  then  proceed  as  in 
Art.  155.. 

23.  The  sum  of  two  numbers  is  42,  and  the  difference  of  their 
squares  is  756  :  what  are  the  numbers?  Ans   12  and  30. 

24.  The -sum  of  two  numbers  is  65,  and  the  diffeierce  of  their 
squares  is  975  :  what  are  the  numbers  ? 


374  CUBE  ROOT.  [SECT.  XVII 

585.  When  the  difference  of  two  numbers  and  the  difference 
of  their  squares  are  given,  to  find  the  numbers. 

.Divide  the  difference  of  the  squares  by  the  difference  of  the 
numbers,  and  the  quotient  will  be  their  sum;  then  proceed  as 
in  Art.  155. 

24.  The  difference  of  two  numbers  is  29,  and  the  difference  rf 
their  squares  is  1885  :  what  are  the  numbers  ? 

EXTRACTION  OP  THE  CUBE  ROOT. 

580.  To  extract  the  cube  root,  is  to  resolve  a  given  number  into 
three  equal  factors  ;  or,  to  find  a  number  which  being  multiplied 
into  itself  twice,  will  produce,  the  given  number.  (Art.  564.) 

Ex.  1.  What  is  the  cube  root  of  64  ? 

Solution. — Resolving  the  given  numbers  into  three  equal  fac- 
tors, we  have  64  =  4X4X4.  Ans.  4. 

2.  What  is  the  cube  root  of  12167  ? 

Operation.  We   first   separate   the 

Col.  i.      Col.  ii.         12167(23      given  number  into  two  pe- 

1st  term    2_       4X2=       8  riods,   by  placing  a  point 

2d     "        4~~     1200  divisor,)  4 167  over  the  units'  figure,  then 

3d     "        63     1389X3=  4167  over  thousands.  This  shows 

us  that  the  root  must  have 

two  figures,  (Art.  562.  Obs.  3,)  and  thus  enables  us  to  find  part 
of  it  at  a  time. 

Beginning  with  the  left  hand  period,  we  find  the  greatest  cube 
of  12  is  8,  the  root  of  which  is  2.  Place  the  2  on  the  right  of 
the  given  number  for  the  first  part  of  the  root,  and  also  in  Col.  I. 
on  the  left  of  the  number.  Multiplying  the  2  into  itself,  write  the 
product  4  in  Col.  II. ;  and  multiplying  4  by  2,  subtract  its  product 
from  the  period,  and  to  the  right  of  the  remainder  bring  down  the 
next  period  for  a  dividend.  Then  adding  2,  the  first  figure  of  the 
root,  to  the  first  term  of  Col.  I.,  and  multiplying  the  sum  by  2, 
we  add  the  product  8  to  the  1st  term  of  Col.  II.,  and  to  this  sum 

Q  JEST.— 586.  Whit  is  it  to  extract  the  cube  root  1 


ARTS.  585-587.]  CUBE  ROOT.  375 

annex  two  ciphers,  for  a  divisor ;  also  add  2,  the  first  figure  of 
the  root,  to  the  2d  term  of  Col.  I.  Finding  the  divisor  is  con- 
tained in  the  dividend  3  times,  we  place  the  3  in  the  root,  also  on 
the  right  of  the  3d  term  of  Col.  I.  Then  multiply  the  3d  terra 
thus  increased,  by  3,  the  figure  last  placed  in  the  root,  and  add 
the .  product  to  the  divisor.  Finally,  we  multiply  this  sum  by  3, 
ant  subtiact  the  product  from  the  dividend.  Ans.  23. 

587,  Hence,  we  derive  the  following  general 

RULE  FOR  EXTRACTING  THE  CUBE  ROOT. 

I.  Separate  the  given  number  into  periods  of  three  figures  each, 
placing  a  point  over  units,  then  over  every  third  figure  towards  the 
left  in  whole  numbers,   and  over  every  third  figure   towards  the 
right  in  decimals. 

II.  Find  the  greatest  cube  in  the  first  period  on  the  left  hand  ; 
place  its  root  on  the  right  of  the  number  for  the  first  figure  of  the 
root,  and  also  in  Col.  I.  on  the  left  of  the  number.      Then  multi- 
plying this  figure  into  itself,  set  the  product  for  the  first  term  in 
Col.  II.  ;  and  multiplying  this  term  by  the  same  figure  again,  sub- 
tract this  product  from  the  period,  and  to  the  remainder  bringdown 
the  next  period  for  a  dividend. 

III.  Adding  the  figure  placed  in  the  root  to  the  first  term  in 
Col.  I.,  multiply  the  sum  by  the  same  figure,  add  the  product  to  the 
first  term,  in  Col.  II.,  and  to  this  sum  annex  two  ciphers,  for  a  di- 
visor ;  also  add  the  figure  of  the  root  to  the  second  term  of  Col.  I. 

IV.  Find  how  many  times  the  divisor  is  contained  in  the  divi- 
dend, and  place  the  result  in  the  root,  and  also  on  the  right  of  the 
third  t</  m  of  Col.  I.    Next  multiply  the  third  term  thus  increased  by 
the  figure  last  placed  in  the  root,  and  add  the  product  to  the  divisor  ; 
then  multiply  this  sum  by  the  same  figure,  and  subtract  the  product 
from  the  dividend.      To  the  remainder  bring  down  the  next  period 
for  a  new  dividend. 

V.  Find  a  new  divisor  in  the  same  manner  that  the  last  divisor 
was  found,  then  divide,  &c.,  as  before  ;  thus  continue  the  operation 
till  the  root  of  all  the  periods  is  found. 

QUEST.— 587.  What  is  the  first  step  in  extracting  the  cube  root?    The  second'    Third t 
Fourth  1    Fifth  1    How  is  the  cube  root  proved  1 


376  CUBE  ROOT.  [SECT.  XVII 

PROOF  —  Multiply  the  root  into  itself  twice,  and  if  the  last  prod- 
uct is  equal  to  the  given  number,  the  work  is  right. 

OBS.  1.  When  there  is  a  remainder,  periods  of  ciphers  may  be  added,  as  in 
square  root. 

2.  If  the  right  hand  period  of  decimals  is  deficient,  this  defi/jency  must  be 
supplied  by  ciphers.  The  root  must  contain  as  many  decimals  as  there  are 
periods  of  decimals  in  the  given  number. 

&§§    Demonstraiion.—rH\m  rule  depends  upon  the  principle  that  the  cube 

f  the  sum  of  two  numbers  is  equal  to  the  cube  of  the  first  part,  added  to  * 

times  the  square  of  the  first  part  into  the  last  part,  added  to  3  times  the  first 

part  into  the  square  of  the  last,  added  to  the  cube  of  the  last  part.     Take  any 

number,  as  23  ;  we  have  23=20-j-3. 


Or,  (23)3=8000+3b-004-540-}-27=12l67. 

.After  subtracting  the  greatest  cube  from  the  left  hand  period,  it  is  plain  the 
remainder  must  contain  3  times  the  square  of  the  first  part  of  the  root  into  the 
last  part,  &c.  Hence,  if  we  divide  the  remainder  by  3  times  the  square  of  the 
first  part  of  the  root,  the  quotient  will  be  the  last  part.  But  it  will  be  seen 
that  the  divisor  is  3  times  the  square  of  the  first  part  of  the  root,  consequently 
the  quotient  must  be  the  last  part  of  the  root  required. 

1.  The  reason  for  separating  the  given  number  into  periods  of  three  figures, 
is  that  the  cube  of  a  number  can  not  have  more  figures  than  triple  the  number 
of  figures  in  the  root,  nor  but  two  less.     It  also  shows  how  many  figures  the  root 
will  contain,  and  thus  enables  us  to  find  part  of  it  at  a  time.  (Art.  562.  Obs.  3.) 

2.  The  reason  for  annexing  2  ciphers  to  the  divisor  is  (manifestly)  because 
the  first  figure  of  the  root,  of  which  the  divisor  is  £  times  the  squape,  stands  in 
tens'  place 

3.  Placing  the  last  figure  of  the  root  on  the  right  of  the  3d  term  in  Col.  I., 
then  multiplying  it  by  this  figure,  and  adding  the  product  to  the  divisor,  and 
this  sum  being  multiplied  by  the  figure  last  placed  in  the  root,  the  product  will 
evidently  be  3  times  the  square  of  the  first  part  of  the  root  into  the  last  part, 
together  with  3  times  the  first  into  the  square  of  the  last  part,  and  the  cube 
of  the  last  part.     In  a  similar  manner  the  operation  may  be  illustrated  in  all 
other  cases. 

Note.  —  The  preceding  method  of  extracting  the  cube  root  was  discovered  by 
the  late  Mr.  Homer  of  Bath,  England,  and  is  often  called  Hornet's  Method, 
(For  the  common  method,  and  its  demonstration  by  cubical  blocks  see  Piicli- 
cal  Arithmetic,  p.  325). 

QUEST.  —  Obs.  When  there  is  a  remainder,  how  proceed  ?  When  the  right  han.l  period 
of  decimals  is  deficient,  what  must  be  done  1  How  many  decimals  must  the  root  contain  ? 
588.  Why  separate  the  given  number  into  periods  of  three  figures  ?  Why  annex  two 
ciphers  to  the  right  of  the  divisor  ? 


ARTS.  588-590.]  CUBE  ROOT.  377 

589.  The  cube  root  of  a  common  fraction  is  found  by  ex- 
tracting the  root  of  its  numerator  and  denominator,  or  by  first 
reducing  it  to  a  decimal. 

A  mixed  number  should  be  reduced  to  an  improper  fraction,  or 
the  fractional  part  to  a  decimal. 

3,  Required  the  cube  root  of  78314.6. 
Operation. 

78314.600(42.784-. 


Col.  I. 

1st  term    4 

Col.  II.                                 78314.1 

16X4  =                     64 

2d      "       8 

4800,  1st  divisor    )     14314 

3d      "     122 

5044X2   =                  10088 

4th     "     124  529200,  2d  divisor)     4226600 

5th     "     1267          538069X7   =  3766483 


6th     "     1274          54698700,  3d  divisor)  460117000 
7th     "     12818       54801244X8   =  438409952 

59O*  When  the  root  is  required  to  many  places  of  decimals, 
the  operation  may  be  contracted  in  the  following  manner. 

First  find  one  more  than  half  the  number  of  decimal  figures  re- 
quired. For  a  new  divisor,  take  as  many  figures  plus  one  on  the 
left  of  the  last  term  in  Col.  II.  as  remain  to  be  found  in  the  root ; 
and  for  a  dividend  retain  one  more  figure  on  tJie  left  of  the  re- 
mainder than  the  divisor  has ;  then  proceed  as  in  the  contraction 
of  division  of  decimals.  (Art.  333.) 

Required  the  cube  root  of  the  following  numbers : 

4.  91125.  8.   10218313.          12.  37.  16.  iff. 

5.  140608.  9.  11543.176.          13.  6.  17.  f^-fr. 

6.  571787.          10.  20.570824.         14.  376.-       18.  44f. 

7.  2515456.        11.  .241804367.        15.  575.         19.  49-^. 

20.  What  is  the  cube  root  of  2  to  eight  decimals  ? 

21.  What  is  the  cube  root  of  -^g-  to  eleven  decimals  ? 

22.  What  is  the  side  of  a  cubical  mound  which  contains  314432 
solid  feet  ? 

Note.  -Similar  solids  are  to  each  other  as  the  cubes  of  their  homologous 
•ides,  or  like  dimensions.     (Leg.  VII.  20,  VIII.  11.  Cor.)     Hence, 

QUEST.— 589.  How  find  the  cube  root  of  a  common  fraction  ?    Of  a  mixed  number  ? 


378  HIGHER    ROOTS.  [SECT.   XVIL 

591*  To  find  the  side  of  a  cube  whose  solidity  shall  be  dou- 
ble, triple,  &c.,  that  of  a  cube  whose  side  is  given. 

Cube  tlw  given  side,  multiply  it  by  the  given  proportion,  ant*  the 
cube  root  of  the  product  will  be  the  side  of  the  cube  required. 

23.  Wk-;t  is  the  side  of  a  cubical  bin  which  contains  8  times  as 
many  solid  feet  as  one  whose  side  is  4  feet  ?  Ans.  8  ft. 

24.  What  is  the  side  of  a  cubical  block  which  contains  4  times 
&s  many  solid  yards  as  one  whose  side  is  6  feet  ? 

25.  If  a  ball  6  inches  in  diameter  weighs  32  Ibs.,  what  is  the 
weight  of  a  ball  whose  diameter  is  3  inches  ? 

26.  If  a  globe  4  ft.  in  diameter  weighs  900  Ibs.,  what  is  the 
weight  of  a  globe  3  ft.  in  diameter? 

592*  To  find  two  mean  proportionals  between  two  given 
numbers. 

Divide  the  greater  number  by  the  less,  and  extract  the  cube  root 
of  the  quotient.  Multiply  the  root  thus  found  by  the  least  of  the 
given  numbers,  and  the  product  will  be  the  least  proportional  sought ; 
then  multiply  the  least  mean  proportional  by  the  same  root,  and  this 
product  will  be  the  greater  mean  proportional  required. 

Find  two  mean  proportionals  between  the  following  numbers : 

27.  8  and  216.  29.   12  and  1500.         31.  71  and  15396. 

28.  64  and  512.        30.  40  and  2560.         32.  83  and  60507. 

EXTRACTION   OP   ROOTS    OF   HIGHER   ORDERS. 

593.  When  the  index  denoting  the  root  to  be  extracted  is  a 
composite  number. 

First  extract  the  root  denoted  by  one  of  the  prime  factors  of  the 
y*4)en  index  /  then  of  this  root  extract  the  root  denoted  by  anotlwr 
f  'ime  factor,  and  so  on.  Thus, 

For  the  4th  root,  extract  the  square  root  twice. 
For  the  6th  root,  extract  the  cube  root  of  the  square  root* 
For  the  8th  root,  extract  the  square  root  three  times. 
For  the.  27th  root,  extract  the  cube  root  three  times. 
1.  What  is  the  4th  root  of  81  ?  Ans.  3. 


ARTS.  591-594.]  HIGHER   ROOTS.  379 

2.  What  is  the  8th  root  of  256  ? 

3.  The  4th  root  of  65530  ? 

4.  The  4th  root  of  19987173376 

5.  The  6th  root  of  46656  ? 

6.  The  6th  root  of  308915776  ? 
7  The  8th  root  of  390625  ? 

•S    The  9th  root  of  40353607  ? 
9    The  18th  root  of  387420489  ? 

10.  The  27th  root  of  134217728? 

594.  When  the  index  denoting  the  root  is  not  a  composite 
number,  we  have  the  following  general 

RULE    FOR    EXTRACTING   ALL    ROOTS. 

1.  Point  off  the  number  into  periods  of  as  many  figures  each,  as 
there  are  units  in  the  given  index,  commencing  with  the  units  figure. 

11.  Find  the  first  f,gure  of  the  root,  and  subtract  its  power  from 
the  left  hand  period ;   then  to  the  right  of  the  remainder  bring  down 
the  first  figure  in  the  next  period  for  a  dividend. 

III.  Involve  the  root  to  the  power  next  inferior  to  that  of  the  index 
of  the  required  root,  and  multiply  it  by  the  index  itself,  for  a  divisor. 

IV.  Find  how  many  times  the  divisor  is  contained  in  the  divi- 
dend, and  the  quotient  will  be  the  next  figure  of  the  root. 

V.  Involve  the  whole  root  to  the  power  denoted  by  the  index  of 
the  required  root,  and  subtract  it  from  the  two  left  hand  periods  of 
fke  given  number. 

VI.  Finally,  bring  down  the  fi.rst  figure  of  the  next  period  to  the 
remainder,  for  a  new  dividend,  and  find  a  new  divisor  as  before 
Thus  proceed  till  the  whole  root  is  extracted. 

OBS.  I.  The  reason  of  this  rule  may  be  illustrated  in  the  same  manner  a« 
thaf,  for  the  extraction  of  the  Square  and  Cube  Roots. 

fc.  The  proof  of  all  roots  is  by  involution. 

3.  Any  root  whatever  may  be  extracted  by  an  extension  of  the  principle  ap- 
plied to  the  extraction  of  the  cube  root.  In  this  general  application  of  the 
principle,  the  given  number  must  be  divided  into  periods,  each  consisting  of  as 
many  figures  as  there  are  units  in  the  index  of  the  required  root,  and  the  num- 
ber of  columns  employed  will  be  one  ess  than  there  are  units  in  the  given  in- 
ilex.  The  operation  then  proceeds  exactly  as  in  the  extraction  of  the  cube 
root ;  and  if  there  be  a  remainder,  a  like  contraction  is  admissible. 
T.H. 


880  HIGHER    ROOTS.  [SECT.   XV11 

4 

11.  Required  the  5th  root  of  35184372088632. 

Operation. 

35184372088832(512  Ans 
3125 
=  3125)  3934 


=  345025251 


51  4X  5  =  33826005) 

512s  =351  84372088832 


12.  Required  the  5th  root  of 

13.*'  Required  the  7th  root  of  2103580000000000000. 

Note.  —  The  preceding  method  in  most  of  the  practical  cases,  gives  ]erhapi 
as  easy  solutions,  as  the  nature  of  the  case  will  admit.  But  when  roots  of  a 
very  high  order  are  required,  the  process  may  be  shortened  by  the  following  * 

APPROXIMATE    RULE. 

595*  Call  the  index  of  the  given  power  n;  and  find  by  trial 
a  number  nearly  equal  to  the  required  root,  and  call  it  the  assumed 
root.  Raise  the  assumed  root  to  the  power  whose  index  is  n.  Then, 

As  n  +  1  time*  this  power,  added  to  n  —  1  times  the  given  number, 
is  to  n  —  1  times  the  same  power  added  to  n-j-1  times  the  yivennum- 
ber,  so  is  the  assumed  root  to  the  true  root  nearly. 

The  number  thus  found  may  le  employed  as  a  new  assumed  root, 
and  tJie  operation  repeated  to  find  a  result  still  nearer  the  true  root. 

14.  Required  the  365th  root  of  1.06. 

Solution.  —  Take  1  for  the  assumed  root,  the  365th  power  of 
which  is  1  ;  and  n  being  365,  we  have  w  +  l  =  366,  and  n  —  1  = 
864.  Then  proceed  in  the  following  manner  : 

1X366  =  366  1X364  =  364 

1.06X364  =  385.84     1.06X366  =  387.96 

As  751.84  :  75L9(F  :  :  1   :  Ans. 

Ans    1.000159b. 

15.  The  7th  root  of  2  ?  17.  The  1  2th  root  of  1.05  ? 

16.  The  9th  root  of  2  ?  18.  The  100th  root  of  100  ? 


*  Button's  Mathematical  Tracts ;  also  Bonnycastle's  Arithmetic* 


ARTS.  595-600.]  PROGRESSION.  381 

SECTION    XVIII. 
PROGRESSION. 

ART.  596*  When  there  is  a  series  of  numbers  such,  that  the 
~atios  of  the  first  to  the  second,  of  the  second  to  the  third,  &c.,  are 
all  equal,  the  numbers  are  said  to  be  in  Continued  Proportion,  or 
Progression.  Progression  is  commonly  divided  into  arithmetical 
and  geometrical. 

Note. — The  terms  arithmetical  and  geometrical  are  used  simply  to  distin- 
guish the  different  kinds  of  progression.  They  both  belong  equally  to  arith- 
metic and  geometry. 

ARITHMETICAL   PROGRESSION. 

5  9  7  •  Numbers  which  increase  or  decrease  by  a  common  differ- 
ence, are  in  arithmetical  progression.  (Art.  474.  Obs.) 

OBS.  1.  Arithmetical  progression  is  sometimes  called  progression  by  difference, 
or  equidijjerent  scries. 

2.  When  the  numbers  increase,  the  series  is  called  ascending;  as,  3,  5,  7,  9, 
11,  &c.  When  they  decrease,  the  series  is  called  descending ;  as,  11, 9,  7,  5,  &c. 

598.  When  four  numbers  are  in  arithmetical  progression  the 
sum  of  the  extremes  is  equal  to  the  sum  of  the  means. 

Thus,  if  5 — 3  =  9—7,  then  will  5  +  7  =  3  +  9. 
Again,  if  three  numbers  are  in  arithmetical  progression,  the  sum 
of  the  extremes  is  double  the  mean. 

Thus,  if  9 — 6=6 — 3,  then  will  9  +  3  =  6  +  6. 

599.  In  any  arithmetical  progression,  the  sum  of  the  two  ex- 
tremes is  equal  to  the  sum  of  any  other  two  terms  equally  distant 
from  the  extremes,  or  equal  to  double  t/ie  middle  term,  when  the 
number  of  terms  is  odd.     Thus,  in  the  series  1,  3,  5,  7,  9,  it  is 
obvious  that  1  +  9  = 


GOO.  In  an  ascending  series,  each  succeeding  term  is  found 
by  adding  the  common  difference  to  the  preceding  term.  Tims, 
if  the  first  term  is  3,  and  the  common  difference  2,  the  series  ia 
3.  5,  7,  9,  11,  13,  15,  19,  17,  21,  &c. 


382  AKTTHMKTTOnu  SECT.  XVIII 


In  a  descending  series,  each  succeeding  term  is  found  by  sub- 
tracting the  common  difference  from  the  preceding  term.  Thus, 
if  the  first  term  is  15,  and  the  common  difference  2,  the  series  u 
15,  13,  11,  9,  7,  &c. 

601.  In  arithmetical  progression  there  are  five  parts  to  be 
considered,  viz  :  the  first  term,  the  last  term,  the  number  of  terms, 
the  common  difference,  and  the  sum  of  all  tlie  terms.     These  parts 
have  such  ar  relation  to  each  other,  that  if  any  three  of  them  are 
given,  the  other  two  may  be  easily  found. 

602.  If  the  sum  of  the  two  extremes  of  an  arithmetical  pro- 
gression is  multiplied  by  the  number  of  the  terms,  the  product 
will  be  double  the  sum  of  all  the  terms  in  the  series. 

Take  the  series  2,     4,     6,     8,  10,  12. 

The  same  inverted  12,   10,     8,     6,     4,     2. 

The  sums  of  the  terms  are  14,   14,   14,   14,   14,  14. 

Thus,  the  sum  of  all  the  terms  in  the  double  series,  is  equal  to 
the  sum  of  the  extremes  repeated  as  many  times  as  there  are  terms  ; 
that  is,  the  sum  of  the  double  series  is  equal  to  12  +  2  multiplied 
by  6.  But  this  is  twice  the  sum  of  the  single  series.  Hence, 

603.  To  find  the  sum  of  all  the  terms,  when  the  extremes 
and  the  number  of  terms  are  given. 

Multiply  half  the  sum  of  the  extremes  by  the  number  of  terms, 
and  the  product  will  be  tJie  sum  of  the  given  series. 

OBS.  The  reason  of  this  process  is  manifest  from  the  preceding  illustration. 

Ex.  1.  The  extremes  of  a  series  are  3  and  25,  and  the  number 
of  terms  is  12  :  what  is  the  sum  of  all  the  terms  ?  Ans.  168. 

2.  What  is  the  sum  of  the  natural  series  of  numbers,  1,  2,  3, 
4,  5,  &c.,  up  to  100? 

3.  Hnw  many  strokes  does  a  common  clock  strike  in  1'2  hours 

GO4«  To  find  the  common  difference,  when  the  extremes  and 
the  number  of  terms  are  given. 

Divide  the  difference  of  the  extremes  by  the  number  of  terms  lea 
1,  and  the  quotient  will  be  the  common  difference  required. 

OBS.  The  truth,  of  this  rule  is  manifest  from  Art.  603. 


ARTS.  601-007. J  PROGRESSION  383 

4.  The  extremes  are  5  and  56,  and  the  number  of  terms  is  18: 
what  is  the  common  difference  ?  Ans.  3. 

5.  If  the  extremes  are  3  and  300,  and  the  number  of  terms  10, 
what  is  the  common  difference  ? 

GO5.  To  find* the  number  of  terms,  when  the  extremes  and 
common  difference  are  given. 

Divide  the  difference  of  the  extremes  by  the  common  difference,  and 
the  quotient  increased  by  1  will  be  the  number  of  terms. 

OBS.  The  truth  of  this  principle  is  manifest  from  the  manner  in  which  the 
successive  terms  of  a  series  are  formed.  (Art.  GOO.) 

6.  If  the  extremes  are  6  and  470,  and  the  common  difference 
is  8,  what  is  the  number  of  terms  ?  Ans.  59. 

7.  If  the  extremes  are  500  and  70,  and  the  common  difference 
is  10,  what  is  the  number  of  terms? 

GOG.  When  the  sum  of  the  series,  the  number  of  terms,  and 
one  of  the  extremes  are  given,  to  find  the  other  extreme. 

Divide  twice  the  sum  of  the  series  by  the  number  of  terms,  and 
from  the  quotient  take  the  given  extreme. 

OBS.  The  reason  of  this  rule  is  manifest  from  Art.  602. 

8.  If  the  sum  of  a  series  is  576,  the  number  of  terms  24,  and 
the  first  term  1,  what  is  the  last  term?  Ans.  47. 

9.  If  the  sum  of  a  series  is  1275,  the  number  of  terms  50,  and 
<he  greater  extreme  47-J-,  what  is  the  less  extreme? 

6O7.  To  find  any  given  term,  when  the  first  term  and  the 
common  difference  are  given. 

Multiply  the  common  difference  by  one  less  than  the  number  of 
terms  required  ;  then  if  the  series  be  ascending,  add  the  product  to 
the  first  term  ;  but  if  it  be  descending,  subtract  it. 

OBS.  The  reason  of  this  rule  may  be  seen  from  the  manner  in  which  th« 
succeeding  terms  of  a  series  are  formed.  (Art.  GOO.) 

10.  If  the  first  term  of  an  ascending  series  is  7,  and  the  common 
difference  3,  what  is  the  41st  term?  Ans.  127. 

11.  If  the  first  term  of  a  descending  series  is  100,  and  the  com- 
mon difference  1-J-,  what  is  til  e  54th  term  ? 


384  GEOMETRICAL  [SECT.   X VIII. 

12.  If  tli.!  first  term  of  an  ascending  series  is  Y,  and  the  com- 
mon difference  5,  what  is  the  100th  term? 

608*  To  find  any  given  number  of  arithmetical  means,  when 
the  extremes  are  given. 

Subtract  tlie  less  extreme  from  the  greater,  and  divide  the  re- 
mainder by  1  more  than  the  number  of  means  required  ;  the  quo- 
tient will  be  the  common  difference,  which  being  continually  added 
to  tlie  less  extreme,  or  subtracted  from  the  greater  extreme,  will  give 
the  mean  terms  required.  One  mean  term  may  be  found  by  taking 
half  the  sum  of  tlie  extremes.  (Art.  598.) 

OBS.  This  rule  depends  upon  the  same  principle  as  that  in  Art.  604. 

13.  Required  3  arithmetical  means  between  7  and  35. 

14.  Required  6  arithmetical  means  between  1  and  99. 

GEOMETRICAL  PROGRESSION. 

6O9»  Numbers  which  increase  by  a  common  multiplier,  01 
decrease  by  a  common  divisor,  are  in  Geometrical  Progression. 

The  numbers  4,  8,  16,  32,  64,  <fec.,  are  in  geometrical  progres- 
sion ;  and  if  each  preceding  term  is  multiplied  by  2,  the  product 
will  be  the  succeeding  term;  thus,  4x2  =  8  ;  8X2  =  16,  etc. 

Again,  if  the  order  of  this  series  be  inverted,  the  proportion 
will  still  be  preserved  and  the  common  multiplier  become  a  com- 
mon divisor.  Thus;  in  the  series  64,  32,  16,  8,  &c.,  64-1-2  =  32; 
32-:-2=16,  &c. 

Note. — If  the  first  term  and  ratio  are  the  same,  the  progression  is  simply 
a  series  of  powers;  as '2;  2x2;  2x2x2;  2x2X2X2,  &c. 

OBS.  1.  Geometrical  Progression  is  geometrical  proportion  continued.  It  is 
therefore  sometimes  called  continual  proportionals,  or  progression  by  quotients. 

If  the  series  increases  it  is  called  ascending;  if  it  decreases,  descending. 

2.  The  numbers  which  form  the  series,  are  called  the  terms  of  the  progres- 
sion.    The  common  multiplier,  or  dirisor,  is  called  the  ral.io.     For  most  pur- 
poses, however,  it  will  .be  more  simple  to  consider  the  ratio  as  always  a  multi- 
plier, either  integral  or  fractional.     Thus,  in  the  series  64,  32,  16,  &c.,  the 
ratio  is  either  2  considered  as  a  divisor,  or  J  considered  as  a  multiplier. 

3.  In  Geometrical  as  well  as  in  Arithmetical  progression,  there  are  five  parts 
to  be  considered,  viz:  the  first  term,  the  last  term,  the  number  of  terms,  the  ratio, 
and  the  sum  of  alt  the  terms.     These  parts  have  such  a  relation  to  each  other, 
that  if  any  three  of  them  are  given,  the  other  two  may  be  easily  found. 


ARTS.  608-011.]  PROGRESSION.  385 

6 1  O.  To  find  the  last  term,  when  the  first  term,  the  ratio, 
and  the  number  of  terms  are  given. 

Multiply  the  first  term  into  that  power  of  ike  rath  ichow  index 
is  1  less  than  the  number  of  terms,  and  the  product  will  be  Ute 
last  terrr-  required. 

OKS.  ]  The  reason  of  this  process  may  be  seen  by  adverting  to  the  mannet 
m  -wh'cK  each  successive  term  is  formed.  (Art.  GOD.)  Thus,  in  the  series  4,  8, 
K;  32,  &c.  the  &l  term  8=4X-):  ll>=4X^X2,  or  4X2-;  fr2=4X23,  &c. 

2.  It  will  be  seen  that  the  several  Amounts  in  compound  interest,  form  a 
gtomctncal  series  of  which  the  principal  is  the  1st  term  ;  the  amount  of  $'l  for 
1  year  the  ratio;  and  the  number  of  years-j-l  the  number  of  terms.  Henco 
the  required  amount,  of  compound  interest  may  be  found  in  the  same  way  as 
the  last  term  of  a  geometrical  series. 

1.  If  the  first  term  of  a  geometrical  progression  is  2,  and  the 
ratio  4,  wh.it  is  the  5th  term?  Art*.  512. 

2.  The  first  term  is  04,  and  the  ratio  •£:  what  is  the  5th  term? 

3.  The  first  term  is  2,  and  the  ratio  3  :  what  is  the  8th  term? 

4.  The  first  term  is  7,  and  the  ratio  5  :  what  is  the  10th  term? 

5.  A  farmer  hired  a  man  for  a  year,  agreeing  to  give  him  &1  for 
the  1st  month,  82  for  the  2d,  &4  for  the  3d,  and  so  on,  doubling 
his  wages  each  month :  how  much  did  he  give  the  last  month? 

G.  What  is  the  amount  of  8250,  at  6  per  cent.,  for  5  years  com- 
pound int.  ?  Of  $500,  at  7  per  ct.,  for  G  years?  Of  $1000,  at 
5  per  ct.,  for  10  years? 

Oil.  To  find  the  sum  of  the  series,  when  the  ratio  and  the 
extremes  are  given. 

Multiply  the  greatest  term  into  the  ratio,  from  the  product  sub- 
tract the  least  term,  and  divide  the  remainder  by  the  ratio  less  1. 

OBS.  1.  When  the  first  term,  the  ratio,  and  the  number  of  terms  are  given,  to 
find  the  sum  of  the  series  we  must  first  find  the  last  term,  then  proceed  us  above. 

2.  The  sum  of  an  infinite  series  whose  terms  -decrease  by  a  common  divisor, 
may  be  found  bit  mitUif^uina'  Die  greatest  term  info  the  ratio,  and  dii't'lri"  Iht 
proffuct  t»i  tJie  ratio  less  1  The  hast,  term  being  infinitely  small,  is  of  no  com- 
parative value,  and  is  therefore  neglected.  $ 

7.  What  is  the  sum  of  the  series,  whose  extremes  are  5  and 
1215,  and  the  ratio  3  ?  Ans.   1820. 

8.  The  extremes  of  a  series  are  1  and  512,  and  the  latio  2* 
what  is  the  sum  of  the  series  ? 


386  ANNUITIES.  [SECT   XVIII. 

9.  The  extremes  of  a  series  are  1024  and  15274^,  and  the  ratio 
is  1£ :  what  is  the  sum  of  the  series  ? 

10.  A  merchant  hired  a  clerk  for  a  year,  and  agreed  to  pay  him 
1  mill  the  1st  month ;  1  cent  the  2d ;  10  cents  the  3d,  and  so  on, 
increasing  in  a  tenfold  ratio  for  each  successive  month  :  what  was 
the  amount  of  his  wages  ? 

11.  What  is  the  sum  of  the  infinite  series  1  +i+-J--f  i,  &c. ;  that 
8,  the  descending  series  whose  first  term  is  1  and  the  ratio  2  V 

Ans.  2. 

12.  What  is  the  sum  of  the  infinite  series  1  +i+i+ izV+'sV,  &c. 

612*  To  find  the  ratio,  when  the  extremes  and  number  of  terms 
are  given. 

Divide  the  greater  extreme  by  the  less,  and  extract  that  root  of  the 
quotient  whose  index  is  1  less  tlmn  the  number  of  terms. 

13.  The  extremes  of  a  series  are  3  and  192,  and  the  number 
of  terms  7  :  what  is  the  ratio  ?  Ans.  2. 

14.  What  is  the  ratio  of  a  series  of  5  terms,  whose  extremes  are 
7  and  567  ? 

Note. — Other  formulas  in  arithmetical  and  geometrical  progression  :might  be 
added,  but  they  involve  principles  with  which  the  student  is  supposed  as  yet 
to  be  unacquainted.  For  a  fuller  discussion  of  the  subject,  see  Thomson' 
Day's  Algebra. 

ANNUITIES. 

(513.  The  term  annuity  properly  signifies  a  sum  of  money 
payable  annually,  for  a  certain  length  of  time,  or  forever. 

OBS.  1.  Payments  made  semi-annually,  quarterly,  monthly,  &c.,  are  also 
called  annuities.  Annuities  therefore  embrace  pensions,  salaries,  rents,  &c. 

2.  When  annuities  remain  unpaid  after  they  are  due,  they  are  said  to  be 
forborne,  or  in  fir  rears.     The  sum  of  the  annuities  in  arrears,  added  to  the  in- 
terest due  on  each,  is  called  the  amount. 

The  present  worth  of  an  annuity  is  the  sum,  which  being  put  at  interest, 
will  exactly  pay  the  annuity. 

3.  When  an  annuity  does  not  commence  till  a  given  time  has  elapsed,  it  is 
called  an  annuity  in  reversion;  when  it  continues/0raw,  a  perpetuity. 

4.  In  finding  the  amount  of  annuities  in  arrears,  it  is  customary  to  reckon 
compound  interest  on  each  annuity  from  the  time  it  is  due  to  the  time  of  pay- 
ment.     The  process  therefore  is  the  same  as  find  ng  th«  sum  of  an  ascending 
geometrical  series.  (Art.  611.)     Hence, 


ARTS   G12-615.J 


ANNUITIES. 


387 


G  1  4.  To  find  the  amount  of  an  annuity  in  arrears. 

Make  the  annuity  t/te  first  term  of  a  geometrical  series,  the 
amount  of  $1  for  1  year  the  ratio,  and  the  given  number  of  years 
t/te  number  of  terms  ;  then  find  the  sum  of  the  series,  and  it  will  be 
the  amount  required.  (Arts.  610,  611.) 

GBS.  When  the  payments  are  not  yearly,  for  the  amount  of  $1  for  1  year,  use 
its  amount  for  the  time  between  the  payments  ;  and  instead  of  the  number  of 
y«ut  s,  use  the  number  of  payments  that  have  been  omitted,  and  proceed  as  before. 

I.  What  is  the  amount  of  an  annuity  of  $100  which  has  not  been  paid  for 
3  years,  at  G  per  cent,  compound  interest  1 

.  06)2=112,36;  and  (112.36X1-06)—  100-*-.06=  $318.36. 


TABLE,  showing  the  amount  of  annuity  of  $1,  or  £1,  at  5,  6,  and  7  per  cent, 
for  any  number  of  years  from  I  to  20. 


Yrs. 

5  |>er  ct. 

6  per  ct.  i  7  per  ct 

Yrs. 

5  per  ct. 

6  per  ct^ 

7  per  ct. 

1 

1.00000 

1.00000 

1.00000 

11 

14.20G78 

14.97164 

15.7836 

2 

2.05000 

2.0GOOO 

2.07000 

12 

15.91712 

16.86994 

17.8884 

3 

3.15250 

3.  18360  1  3.21490 

13 

17.71298 

18.88213 

20.1406 

4 

4.31012 

4.37461  1443994: 

14 

19.598G3 

2  1.0  1506  122.5504 

5 

5.52563 

5.G3709  |  5.75073  ; 

15 

21.5785G 

23.27596  1  25.  1290 

G 

6.80191 

G.97532 

7.15329 

16 

23.65741) 

25.67252  i  27.8880 

7 

8.14201 

8.39383  8.65402 

17 

25.84036 

28.21287 

30.8402 

8 

9.541)11 

9.8974G 

10.2598 

18 

28.13238 

30.90565 

33.9990 

1) 

1  1  .02656 

11.49131 

11.9799 

19 

30.53900 

33.75999 

37.3789 

10 

12.57789 

13.18079 

13.8164 

20  ;  33.0G595  j  3G.78559  j  40.9954 

Note. — Multiply  the  given  annuity  by  the  amt.  of  $1,  for  the  given  number 
of  years  found  in  the  Table,  and  the  product  will  be  the  amount  required. 

3.  What  will  an  annual  rent  of  $75  amount  to  in  9  years,  at  5  per  cent.  1 

4.  What  is  the  amount  of  $200  forborne  for  9  years,  at  6  per  cent.  1 

5.  What  is  the  amount  of  $350  forborne  for  10  years,  at  7  per  cent.  ? 

6.  What  is  the  amount  of  $1000  forborne  for  20  years,  at  6  per  cent.  ? 

615*  To  find  the  present  worth  of  an  annuity. 

Find  the  amount  of  $1  annuity  for  the  given  time  as  before  ; 
then  divide  this  amount  by  the  amount  of  $1  at  compound,  interest 
for  the  same  time,  multiply  the  quotient  by  the  given  annuity,  and 
the  product  will  be  the  present  ivorth.  If  the  annuity  is  a  perpetuity, 
or  to  continue  forever,  multiply  it  by  100,  divide  the  product  by 
the  given  rate,  and  the  quotient  ivill  be  the  present  value  required. 

OBS.  For  the  amount  of  $1  at  compound  interest,  see  Table,  p.  271. 

7.  What  is  the  present  worth  of  an  annuity  of  $40  to  continue  5  years,  at 
5  per  cent,  compound  interest"?  Ans.  $173.178. 

17* 


388  PERMUTATIONS.  [SECT.    XVIII. 

8.  What  is  the  present  worth  of  an  annuity  of  $80  to  continue  forever,  at 
6  per  cent.  1 

6 1  G»  To  find  the  present  worth  of  an  annuity  in  reversion. 

Find  the  present  worth  of  the  annuity  from  the  present  time  till 
its  termination  ;  also  find  its  present  worth  for  the  time  before  it 
commences  ;  the  difference  between  these  two  results  will  be  tlie  pres- 
ent worth  required. 

9.  What  is  the  present  worth  of  $79.625  at  5  per  cent.,  to  commence  in 
years  tJid  continue  6  years  1  Ans.  $332.50. 

PERMUTATIONS    AND    COMBINATIONS. 

617.  By  Permutations  is  meant  the  changes  which  may  be 
made  in  the  arrangement  of  any  given  number  of  things. 

The  tern,  combinations,  denotes  the  taking  of  a  less  number  of 
things  out  of  a  greater,  without  regard  to  their  order  or  position. 

618.  To  find  how  many  permutations  or  changes  may  be  made 
in  the  arrangement  of  any  given  number  of  things. 

Multiply  together  all  the  terms  of  the  natural  series  of  numbers 
from  1  up  to  the  given  number,  and  the  product  will  be  the  answer. 

1.  How  many  changes  may  be  rung  on  5  bells!  Ans.  120. 

2.  How  many  diff&rent  ways  may  a  class  of  8  pupils  be  arranged  1 

3.  How  many  different  ways  may  a  family  of  9  children  be  seated  1 

4.  How  many  ways  may  the  letters  in  the  word  arithmetic,  be  arranged? 

5.  A  club  of  12  persons  agreed  to  dine  with  a  landlord  as  long  as  he  could 
seat  them  differently  at  the  table  :  how  long  did  their  engagement  last? 

6 1 9»  To  find  how  many  combinations  may  be  made  out  of 
any  given  number  of  different  things  by  taking  a  given  number 
of  them  at  a  time. 

Tak$  the  series  of  numbers,  beginning  at  the  number  of  things 
given,  and  decreasing  by  1  till  the  number  of  terms  is  equal  to  the 
number  of  things  taken  at  a  time ;  the  product  of  all  the  terms 
\4>iil  be  the  answer  required. 

6.  How  many  different  words  can  be  formed  of  9  letters,  taking  3  at  a  time  1 
Solution.— 9X8X7=504.  Ans.  504  words. 

7-  Fow  many  numbers  can  be  expressed  by  the  9  digits,  taking  5  at  a  time  1 
Q.  Mow  many  words  of  fi  letters  each  can  be  formed  out  of  the  2b'  letters  of 
lie  alphabet,  on  the  supposition  t\at  consonants  will  form  a  word  1 


ARTS.  610-025.]  MENSURATION.  389 

^- 

SECTION    XIX. 
APPLICATION  OF  ARITHMETIC  TO  GEOMETRY. 

G2O.  In  the  preceding  sections  abstract  numbers  have  been 
applied  to  concrete  substances,  or  to  objects  in  general,  considered 
arithmetically.  On  the  same  principle,  geometrical  magnitudes 
aiay  be  ccmpared  or  measured  by  means  of  the  numbers  repre- 
senting their  dimensions.  (Arts.  7,  510.  Oba.  3.) 

Ous    The  measurement  of  magnitudes  is  commonly  called  mensuration. 

MENSURATION   OF   SURFACES. 

G2  I  •  In  the  measurement  of  surfaces,  it  is  customary  to  assume 
a  square  as  the  measuring  unit,  whose  side  is  a  linear  unit  of 
the  name  name.  (Leg.  IV.  4.  Sch.  Art.  257.  Obs.  2.) 

Note. — For  the  demonstration  of  the  following  principles,  see  references. 

G22.  To  find  the  area  of  a  parallelogram,  also  of  a  square." 
Multiply  the  Icnyth  l)>j  the  breadth.  (Art.  285,  Leg.  IV.  5.) 

OBS.  When  the  area  and  one  side  of  a  rectangle  are  given,  the  other  side  it 
Rrnnd  by  dividing  the  area  by  the  given  side.  (Art.  15G.) 

1.   How  many  acres  in  a  field  '240  rods  long,  and  180  rods  wide! 

*«J.  How  many  acres  in  a  square  field  the  length  of  whose  side  is  3-10  rods! 

3.  If  the  diagonal  of  a  square  is  100  rods,  what  is  its  urea 7 

4.  A  rectangular  farm  of  320  acres,  is  £  a  mile  wide:  what  is  its  length! 

G23.  To  find  the  area  of  a  rhombus.  (Leg.  I.  Def.  18.  IV.  5  ) 
Multiply  the  Icnyth  by  the  altitude  or  perpendicular  height. 

5.  Find  the  area  of  a  rhombus  whose  length  is  20  ft.,  and  its  altitude  18  ft 
G24.  To  find  the  area  of  a  trapezium.  (Leg.  IV.  7.) 
Multiply  half  the  sum  of  the  parallel  sides  by  the  altitude. 

6.  Find  the  area  of  a  trapezium  the  lengths  of  whose  parallel  sides  art 
27  ft.  and  31  it,  and  whose  altitude  is  15  it. 

G25.  To  find  the  area  of  a  triangle. .  (Leg.  IV.  C.) 
Multiply  the  base  by  half  the  altitude  or  perpendicular  hejyht. 

7.  Find  the  area  of  a  triangle  whose  base  is  50  't.}  end  its  altitude  44  ft. 


390  MENSURATION.  |  SECT.  XIX. 

626*  To  find  the  area  of  a  triangle,  the  thrpe  sides  being  given. 

From  half  the  sum  of  the  three  sides  subtract  each  side  respec- 
tively ;  then  multiply  together  half  the  sum  and  'he  three  remain- 
ders, and  extract  the  square  root  of  the  product. 

9.  What  is  the  area  of  a  triangle  whose  sides  are  20,  30,  and  10  ft.  I 

10.  How  many  acres  in  a  triangle  whose  sides  are  each  40  rods  1 

627.  To  find  the  circumference  of  a  circle  from  its  diameter. 
Multiply  the  diameter  by  3.14159.  (Leg.  V.  11.  Sch.) 

Nole. — The  circumference  of  a  circle  is  a  curve  line,  all  the  points  of  which 
are  equally  distant  from  a  point  within,  called  the  centre.  The  diameter  cf  a 
circle  is  a  straight  line  which  passes  through  the  centre,  and  is  terminated  on 
both  sides  by  the  circumference.  The  radius  or  semi-diameter  is  a  straight 
line  drawn  from  the  centre  to  the  circumference, 

11.  What  is  the  circumference  of  a  circle,  whose  diameter  is  20  ft.  7 

12.  What  is  the  circumference  of  a  circle,  whose  diameter  is  45  rods  1 

G28.  To  find  the  diameter  of  a  circle  from  its  circumference. 
Divide  the  circumference  by  3.14159. 

OBS.  The  diameter  of  a  circle  may  also  be  found  by  dividing  the  area  by 
7854,  and  extracting  the  square  root  of  the  quotient. 

13.  What  is  the  diameter  of  a  circle,  whose  circumference  is  314.159  ft.  1 

629.  To  find  the  area  of  a  circle.  (Leg.  V.  11.) 

Multiply  half  the  circumference  by  half  the  diameter  ;  or,  mul- 
tiply the  square  of  the  diameter  by  the  decimal  .7854. 

15.  What  is  the  area  of  a  circle,  whose  diameter  is  50  rods  ? 

l(j.  Find  the  area  of  a  circle  200  ft.  in  diameter,  and  628.318  ft.  in  circura. 

630.  To  find  the  side  of  the  greatest  square  that  can  be  in- 
scribed in  a  circle  of  a  given  diameter. 

Divide  the  square  of  the  given  diameter  by  2,  and  extract  the 
square  root  of  the  quotient.  (Art.  581.  Obs.  1.) 

17.  The  diameter  of  a  round  table  is  4  ft. ;  what  is  the  side  of-the  greatest 
square  table  which  can  be  made  from  it  7 

63  I.  To  find  the  side  of  the  greatest  equilateral  triangle  that 
Can  be  inscribed  in  a  circle  of  a  given  diameter. 

Multiply  $  the  given  diameter  by  1.73205.  (Leg.  V.  4.  Sch.) 

18.  Required  the  side  of  an  equilateral  triangle  inscribed  in  a  circle  of  20  ft, 
<liametcr, 


ARTS.  626-637.]  MENSURATION.  391 

MEASUREMENT   OF    SOLIDS. 

632.  In  the  measurement  of  solids  it  is  customary  to  assume 
a  cube  as  the  measuring  unit,  whose  sides  are  squares  of  the  same 
name.  (Art.  258.  Obs.  2.) 

633.  To  find  the  solidity  of  bodies  whose  sides  are  perpen- 
dicular to  each  other. 

Multiph   the  length,  breadth,  and  thickness  together.  (Art.  286.) 

OBS.  When  the  contents  of  a  solid  body  and  two  of  its  sides  are  given,  the 
tftker  side  is  found  by  dividing  the  contents  by  the  product  of  the  two  given 
sides.  (Art.  159.) 

1.  What  are  the  contents  of  a  stick  of  timber  4  ft.  square,  and  85J  ft.  long! 

2.  What  is  the  capacity  of  a  cubical  vessel,  14  ft.  8  in.  deep  1 

634.  To  find  the  solidity  of  a  prism. 

Multiply  the  area  of  the  base  by  the  height.  (Leg.  VII.  12.) 

OBS.  This  rule  is  applicable  to  all  prisms,  triangular,  quadrangular,  pentag- 
onal, &c.,  also  to  all  parallelopipedons^  whether  rectangular  or  oblique. 

3.  Find  the  solidity  of  a  prism  4G£  ft.  high,  whose  base  is  7£  ft.  square  1 

635.  To  find  the  lateral  surface  of  a  right  prism. 
Multiply  the  length  by  the  'perimeter  of  its  base.   (Leg.  VII.  5.) 

OBS.  If  we  add  the  areas  of  both  ends  to  the  lateral  surface,  the  sum  will  be 
the  whole  surface  of  the  prism. 

4.  What  is  the  surface  of  a  triangular  prism,  whose  sides  are  each  3  ft.,  and 
its  length  12  ft.  ? 

636.  To  find  the  solidity  of  a  pyramid  and  cone. 
Multiply  the  area  of  the  base  by  \  of  the  height.   (Leg.  VII.  18.) 

5.  What  is  the  solidity  of  a  pyramid  100  ft.  high,  whose  base  is  40  ft.  square  1 

6.  What  is  the  solidity  of  a  cone  150  ft.  high,  whose  base  is  15  ft.  in  diameter  1 

63  7.  To  find  the  lateral  or  convex  surface  of  a  regular  pyra- 
mid, or  cone.  (Leg.  VII.  1G,  VIII.  3.) 

Multiply  the  perimeter  of  the  base  by  %  the  slant-height. 

7.  What  is  the  lateral  surface  of  a  regular  pyramid,  whose  slant-height  is, 
15  ft.,  and  base  is  30  ft.  square  1 

8.  What  is  the  convex  surface  of  a  right  cr  ne,  whose  slant-height  is  94  ft. 
and  the  perimeter  of  its  base  37  ft. '{ 


393  MENSURATION.  [SECT.    XIX 

638.  To  find  the  solidity  of  a  frustum  of  a  pyramid  and  cone. 
To  the  sum  of  t/ie  areas  of  the  tivo  enda,  add  the  square  root  of 

the  product  of  these  areas  ;  then  multiply  thin  sum  by  %  of  ike  per- 
pendicular heiyht.  (Leg.  VII.  19.  Sch.,  Vlll.G.) 

9.  If  the  two  ends  of  the  frustum  of  a  pyramid  are  3  ft  and  2  ft.  square,  and 
the  height  is  12  ft.,  what  is  its  solidity  1 

639.  The  convex  surface  of  a  frustum  of  a  pyramid  and  cone 
is  found  by  multiplying  half  the  sum  of  the  circumferences  of  tto 
to;>  ends  by  the  slant-height.  (Leg.  VII.  17,  V11I.  5.) 

10.  If  the  circumferences  of  the  two  ends  of  the  frustum  of  a  cone  are  18  ft 
and  14  ft.,  and  its  slant-height  11  ft.,  what  is  its  convex  surface  1 

640.  To  find  the  solidity  of  a  cylinder. 

Multiply  the  area  of  the  base  by  the  heir/hi.  (Leg.  VIII.  2.) 

11.  Find  the  solidity  of  a  cylinder  10  ft.  in  diameter,  and  35  ft.  high. 

12.  Find  the  solidity  of  a  cylinder  100  ft.  in  circumference,  and  150  ft.  high- 

641.  To  find  the  convex  surface  of  a  cylinder. 

Multiply  the  circumference  of  the  base  by  the  heiyht.  (Leg.  VIII.  1.) 

13.  Find  the  convex  surface  of  a  cylinder  5  yds.  in  diameter,  and  5  yds.  long. 

642.  To  find  the  convex  surface  of  a  sphere  or  globe. 
Multiply  the  circumference  by  the  diameter.  (Leg.  VIII.  9.) 

14.  What  is  the  surface  of  a  globe  18  inches  in  diameter  ? 

15.  If  the  diameter  of  the  moon  is  2162  miles,  what  is  its  surface1? 

643.  To  find  the  solidity  of  a  sphere  or  globe. 
Multiply  the  surface  by  •§•  of  the  diameter.   (Leg.  VIII.  11.) 

Hi.   Find  the  solidity  of  a  globe  15  inches  in  diameter. 

17.  The  diameter  of  the  moon  is  21(52  miles:  what  is  its  solidity  7 

MEASUREMENT   OF   LUMBER. 

C44.  The  area  of  a  board  is  found  by  multiplying  the  length  into  the  men* 
tread  A.  (Arts.  622,023.) 

Tr  ?  solid  contents  of  hewn  or  square  timber  are  found  by  multiplying  lh« 
wnsih  into  the  mean  breadth  and  depth. 

The  solid  contents  of  round  timber  are  found  by  multiplying  the  length 
by  \  Uie  mean  girt  or  circumference. 

Obs.  1.  The  mean  breadth  of  a  tapering  board  is  found  by  measuring  i'  in 
the  middle,  >r  by  taking  |  the  sum  of  the  breadths  of  the  two  ends. 


ARTS.  638-647.]  MENSURATION.  3113 

2.  The  mean  dimensions  of  square  and  round  timber  are  found  in  a  similar 
manner. 

3.  The  method  for  finding  the  solidity  of  round  timber  makes  an  allowance 
of  about  -£  tor  waste  in  hewing.  (Arts.  640,  258.  Obs.  3.) 

18  Find  the  area  of  a  board  12  ft.  long,  and  the  ends  14  in.  and  12  in.  wide. 

19  Find  the  solidity  of  a  joist  10  tl.  long,  the  ends  being  8  in.  and  4  in.  sq. 
2(1.   Find  the  solidity  of  a  log  50  ft.  long,  the  circumferences  of  the  ends 

oeing  6  ft.  and  4  ft. 

GAUGING   OF   CASKS. 

645.  The  process  of  finding  the  contents  or  capacities  of  casks 
and  other  vessels  is  called  GAUGING. 

646.  rrh&  contents  of  casks  are  found  by  multiplying  the  square  of  the  mean 
diameter  into  the  length;  then  this  product  multiplied  bi/  .0034  will  give  the  wine 
gallons,  and  multiplied  by  .0028  will  give  the  beer  gallons. 

OES.  The  mean  diameter  of  a  cask  is  found  by  adding  to  the  head  diameter 
.7  of  the  difference  between  the  head  and  bung  diameters  when  the  staves  are 
very  much  curved ;  or  by  adding  .5  when  very  little  curved ;  and  by  adding  .55 
when  they  are  of  a  medium  curve. 

21.  How  many  wine  gallons  in  a  cask  but  little  curved,  whose  length  is  4.5 
in.,  its  bung  diameter  40  in.,  and  its  head  diameter  3G  in.  1 

22.  How  many  beer  gallons  in  a  cask  much  curved,  whose  length  is  G4  in., 
its  bung  diameter  52  in.,  and  head  diameter  40  in.  1 

TONNAGE   OF    VESSELS. 

647.  Government  Rule.— I.  If  the  vessel  be  double-decked,  take  the  lengln 
from  the  fore  part  of  the  main  stern  to  the  after  part  of  the  stern-post,  above  the 
upper  deck  ;  then  the  breadth  at  the  broadest  part  above  the  main  wales,  half 
of  which  breadth  shall  be  accounted  the  depth  of  such  vessel;  from  the  length 
deduct  throe-filths  of  the  breadth,  multiply  the  remainder  by  the  breadth  and 
the  product  by  the  depth  ;  divide  the  last  product  by  95,  and  the  quotient  shall 
be  deemed  the  true  tonnage  of  the  vessel. 

II.  If  the  vessel  be  single-decked,  take  the  length  and  breadth  as  above  di- 
rected, deduct  from  the  length  three-fifths  of  the  breadth,  and  take  the  depth 
from  the  under  side  of  the  deck  plank  to  the  ceiling  in  the  hold,  then  multiply 
and  divide  as  before,  and  the  quotient  shall  be  deemed  the  tonnage. 

Carpenter's  Rule. — The  continued  product  of  the  length  of  the  keel,  the 
brea  1th  at  the  main  beam,  and  the  depth  of  the  hold  in  feet,  divided  by  95  will 
give  the  tonnage  of  a  single-decked  vessel.  For  a  doabl-  .decker^  instead  of 
the  depth  of  the  hold,  take  half  the  breadth  at  the  beam. 

23.  What  is  the  government  tonnage  of  a  dr  uble-decker  whose  engtl  k 
150  ft.,  the  breadth  35  ft.,  and  the  depth  25  ft.  1 

24.  What  is  the  carpenter's  tonnage  of  the  same  vessell 


394  MECHANICAL    POWERS.  [SECT.  XIX. 

MECHANICAL     POWERS. 

G48«  The  Mechanical  powers  are  six,  viz:  the  lever,  the  wlieel 
and  axle,  the  pulley,  the  inclined  plane,  the  screw,  and  the  wedge. 

649.  When  the  power  and  weight  act  perpendicularly  to  the  arms  of  a 
straight  fcccr,  the  power  is  to  the  weight,  as  the  distance  from  die  fulcrum  to 
the  wcig/U  is  to  the  distance  from  the  fulcrum  to  the  power. 

1.  If  the  power  is  100  Ibs.,  the  long  arm  10  ft.,  and  the  short  arm  2  ft.,  what 
weight  can  be  ra'.sed  1 

2.  The  arms  of  a  lever  are  15  ft.  and  4  ft.,  and  the  weight  raised  500  Ibs.: 
«rhat  is  the  lower  7 

650.  When  a  weight  is  sustained  by  a  lever  resting  on  two  props, 

Tke  long  arm  :  the  short  arm  : :  the  weight,  supported  by  the  long  arm  :  the 
weight  supported  by  the  short  arm.     Hence, 

The  whole  length  :  short  arm  : :  whole  weight :  weight  on  s.  a.  (Leg.  III.  16.) 

3.  A  and  B  carry  25G  Ibs.  suspended  upon  a  pole  5  ft.  from  A  and  3  ft.  from 
B  :  how  many  pounds  does  each  carry  1 

4.  A  and  B  carry  90  Ibs.  upon  a  lever  1*2  ft.  long  :  where  must  it  be  placed, 
that  B  may  carry  £  of  it  1 

651.  The  wheel  and  axle  operate  on  the  same  principle  as  the  lercr;  the 
semi-diameter  of  the  wheel  answers  to  the  long  arm,  and  the  semi-diameter  of 
the  axle  to  the  short  arm. 

5.  If  the  diameter  of  a  wheel  is  6  ft.,  and  that  of  the  axle  1  ft.,  what  weight 
will  100  Ibs.  raise  1 

6.  A  wheel  is  8  ft.  diameter,  an  axle  1£  ft. :  what  weight  will  200  Ibs.  raise  1 

652.  In  the  application  of  movable  pulleys, 

The  POWER  :  the  WEIGHT  : :  1  :  twice  l/ie  NUMBER  of  pulleys. 

7.  What  weight  can  a  power  of  200  Ibs.  raise  with  4  movable  pulleys? 

8.  What  power  with  8  pulleys  will  raise  a  pillar  of  granite  weighing  10  tono  • 

653.  The  perpendicular  height  of  an  inclined  plane  is  to  its  length,  as  the 
power  to  the  weight. 

9.  What  power  will  draw  a  train  of  cars  weighing  100000  pounds  up  an  in- 
clined plane  which  rises  GO  ft.  to  a  mile  1 

654.  The  scrsw  acts  upon  the  principle  of  the  inclined  plane      Henc* , 
The  distance  between  the  threads  is  to  the  circumference  of  a  circle  described  toy 

the  pmver,  as  the  power  is  to  the  weight. 

10.  What  weight  can  be  raised  by  a  pr  wer  of  1000  Ibs.  applied  to  a  screw 
whose  threads  are  1  inch  apart,  at  the  end  of  a  lever  12  ft.  long  7 

655.  The  powr-  applied  to  the  head  of  a  wedge  is  to  the  weight,  as  half  t.he 
thickness  of  the  uead  is  to  the  length  jf  its  side.     In  the  use  of  the  w;dge,  cot 
less  than  half  the  power  is  lost  by  fr.ction  against  the  si  les. 


MISCELLANEOUS    SAMPLES.  395 

MISCELLANEOUS   EXAMPLES. 

1.  The  sum  of  two  numbers  is  980,  and  their  difference  62:  what  are  the 
numbers  1 

2.  The  product  of  two  numbers  is  4410,  and  one  is  63  :  what  is  the  other  1 

3.  What  number  multiplied  by  28^-,  will  produce  1451 

4.  What  number  multiplied  by  6£,  will  be  equal  to  7|  multiplied  by  54  } 

5.  If  an  armyof  24000  men  have  520000  Ibs.  of  bread,  how  long  will  it .  *tt 
them,  allowing  each  man  1£  Ibs.  per  day  1 

6.  What  is  the  interest  of  $5256  for  60  days,  at  7  per  cent.  7 

7.  What  is  the  amount  of  $16230  for  4  months,  at  6£  per  cent.  ? 

8.  What  is  the  bank  discount  on  $1200  for  90  days,  at  6  per  cent  1 

9.  For  what  sum  must  a  note  be  made,  payable  in  4  months,  the  proceed* 
of  which  shall  be  $1800,  discounted  at  a  bank  at  7  per  cent.  1 

10.  A  capitalist  sent  a  broker  $25000  to  invest  in  cotton,  after  deducting  hi* 
commission  of  2£  per  cent. :  what  amount  of  cotton  ought  he  to  receive  1 

11.  A  merchant  bought  500  yards  of  cloth  for  $1800:  how  must  he  retail  it 
by  the  yard  to  gain  25  per  cent.  1 

12.  A  man  bought  640  bbls.  of  beef  for  $5000,  and  sold  it  at  a  loss  of  12 
per  cent. :  how  much  did  he  get  a  barrel  1 

13.  If  a  man  buys  1000  geographies,  at  37  J  cents  apiece,  and  retails  them 
at  50  cents,  what  per  cent,  will  he  make  1 

14.  A  grocer  bought  180  boxes  of  lemons  for  $360,  and  sold  them  at  10  per 
cent,  less  than  cost :  what  did  he  lose  1 

15.  How  many  dollars,  each  weighing  412£  grs.,  can  be  made  from  16  Ibs. 
5  oz.  of  silver  1 

16.  How  many  eagles,  weighing  258  grs.  apiece,  will  21  Ibs.  10  oz.  make! 

17.  How  long  a  thread  can  be  spun  from  1  ton  of  flax,  allowing  5  oz.  will 
make  100  rods  of  thread  1 

18.  How  many  revolutions  will  the  hind  wheel  of  a  carriage  5  ft.  6  in.  in 
circumference,  make  in  2  miles  4  furlongs? 

19.  How  many  revolutions  will  the  fore  wheel  of  a  carriage  4  ft.  7  in.  in 
circumference,  make  in  the  same  distance  1 

20.  'Bought  1500  doz.  buttons  for  $187.50:  what  was  that  per  gross  1 

21.  A  man  paid  $132  for  40  bbls.  of  cider:  what  is  that  a  quart  1 

22.  A  man  paid  $150  for  10  rods  of  land,  what  was  that  per  acre  1 

23.  A  man  having  $2500,  laid  out  -£  of  it  in  flour,  at  $5  per  barrel :  how 
many  barrels  did  he  buy  1 

24.  The  commander  of  an  exploring  expedition  found  that  ^  of  his  provision* 
were  exhausted  in  28  months  :  how  much  longer  would  they  last  1 

25    What  cost  15-f  Ibs.  of  cheese,  at  $8£  per  hundred  7 

26.  How  many  yards  of  carpeting  \  yd.  wide  will  it  take  to  cover  a  floor 
18  ft.  long  and  15  ft.  wide  1 

27.  If  ^  yard  of  calico  cost  -f^s.,  what  will  •£  of  an  ell  English  cost? 

28.  How  long  will  468256  Ibs.  of  beef  last  an  army  of  *^45  soldiers,  allow- 
ing each  man  1J  lb.  per  day  1 


398  MISCELLANEOUS    EXAMPLES. 

29.  How  long  would  the  same  quantity  of  beef  last  Jhe  army,  if  reinforced 
jy  2500  men,  allowing  each  man  1 J  Ib.  per  Jay  '{ 

30.  Bought  f  of  a  pipe  of  wine  for  $120:  what  was  that  per  gallon  7 

31.  If  a  man  can  walk  17  miles  in  5  hours,  12  minutes,  31  seconds,  how 
far  <*an  he  walk  in  3  hours,  40  minutes,  3u'  seconds'? 

32.  If  a  man  traveling  1 1  hours  per  day,  performs  half  his  journey  in  9 
days,  how  long  will  it  take  him  to  go  the  other  half  traveling  10  hours  a  day! 

33.  li  you  lend  a  man  $700  for  99  days,  how  long  ought  he  to  lend  you 
ftll'OO  to  requite  the  favor  1 

34.  A  milkman's  measure  was  deficient  half  a  gill  to  a  gallon :  how  much 
did  he  cheat  his  customers  in  selling  87*20  gallons  7 

35.  If  85  yds.  of  calico  cost  $10/20,  what  will  1500  yds.  cost  1 
3G.  If  1G50  Ibs.  of  sugar  cost  $20:125,  what  will  87  Ibs.  cost! 

37.  If  14-24  gals,  of  oil  cost  $10b'2,  what  will  '210  gals,  cost  7 

38.  If  wind  moves  2$  miles  per  hour,  how  long  is  it  in  moving  from  the  pole 
to  the  equator,  a  distance  of  G214  miles'? 

3D.  If  -j^-  of  a  barrel  of  flour  costs  f  of  a  dollar,  what  will  £  of  a  bbl.  cost  1 

40.  If  £  of  a  ton  of  chalk  cost  £^f,  what  will  -£  of  a  ton  cost  1 

41.  If  |-  of  a  bushel  of  wheat  cost  $|-,  how  much  will  f-  of  a  bushel  cost  1 
4'2.  If  £  of  a  ship  cost  $10000,  what  will  -fa  of  her  cost  1 

43.  If  IGi  bbls.  of  mackerel  cost  $b'5f,  what  will  48^  bbls.  cost  7 

44.  If  28i  gals,  of  oil  cost  $31.25,  how  much  will  $250  buy  7 

45.  At  $&£  for  40-doz.  eggs,  what  will  4b'0-£-  doz.  cost  7 

4G.  The  President's  salary  is  $25000  per  annum :  how  much  can  he  spend 
per  day,  and  lay  up  $10000  of  it  7 

47.  If  one  person  lies  in  bed  9  hours  per  day,  and  another  G  hours,  how 
much  time  will  the  one  gain  over  the  other  in  '20  years  7 

48.  A  cistern  has  three  faucets ;  the  1st  will  empty  it  in  10  min.,  the  2d  in 
20  min.,  the  3d  in  30  min. :  in  what  time  will  they  all  empty  it  7 

4!).  A  man  and  his  wife  drink  a  barrel  of  beer  in  30  days,  and  the  man  alone 
can  drink  it  in  40  days :  how  long  will  it  last  the  wife  7 

50.  A  teacher  being  asked  how  many  scholars  he  had,  replied  \  study  Arith- 
metic, -i-  study  Latin,  -fo  study  Algebra,  -^  study  Geometry,  and  24  study 
French :  how  many  scholars  had  he  7 

51.  A  man  having  spent  i  and  £  of  of  his  money,  had  £48j  left:  how  much 
had  he  at  first  7 

52.  A  man  bequeathed  -^   -f  his  property  to  his  wife,  ±  to  his  son,  -J-  to  hig 
daughter,  and  the  remaindc/,  which  was  $1500,  to  the  Bible  Society:  what 
did  bis  whole  property  amount  to  7 

53    What  is  that  number  •£  of  which  exceeds  -|-  of  it  by  45  7 

54.  Pewter  is  composed  of  1 12  parts  of  tin.  15  of  lead,  and  G  of  brass:  hew 
much  will  it  take  of  each  ingredient  to  make  5(>50  pounds  of  pewter  7 

55.  Two  travelers  start  at  thn  same  time  from  Boston  and  Washington  to 
meet  each  other;  one  goes  5  miles  an  hour,  the  other  7  miles;  the  whole  dis- 
tance is  436  miles :  how  far  will  each  travel  7 


MISCELLANEOUS    EXAMPLES.  397 

56.  A  grocer  divided  ,a  barrel  of  flour  into  2  part?,  so  that  the  smaller  con 
taincd  I  as  much  as  the  other:  how  many  pounds  were  there  in  each  7 

57.  A,  B,  and  C,  build  a  ship  together;  A  advanced  $1000.  B  $12000,  and 
C  $13000;  they  gain  $5000:  what  is  the  gain  of  each? 

58.  A,  H,  and  (J,  entered  into  partnership  ;   A  furnished  $000,  R  and  C  to- 
gether $lb<10  ;  they  gained  $%0,  of  which  B  took  $280  ;  how  much  did  A  and 
C  gain  ;  and  B  and  C  put  in  respectively  I 

51).  The  liabilities  of  a  bankrupt  are  $03240,  and  his  assets  $12018:  wha 
per  cent,  can  he  pay  7 

00.  A  bankrupt  compromises  with  his  creditors  for  37£  per  cent.:  h<  w  mucB 
will  he  pay  on  a  claim  of  $30507 

01.  How  much  will  he  pay  on  a  debt  of  SI  2080 .375  7 

02.  A  owns  f  and  B  -fa  of  a  ship  ;  A's  share  is  worth  $10000  more  thai. 
B's:  what  is  the  value  of  the  ship  7 

03.  A  man  gave  his  oldest  son  -£  of  hi*  t  ropcrty  less  $50;  to  the  second,  he 
gave  •£ ;  and  to  the  youngest  he  gave  tl  >  remainder,  which  was  •£•  less  $10 : 
what  was  the  amount  of  his  property! 

01.  A  man  and  boy  together  can  frame  a  house  in  9  days;  the  man  can 
frame  it  alone  in  1*2  days:  how  long  will  it  take  the  boy  to  frame  it  7 

05.  A  cistern  has  a  receiving  and  a  discharging  pipe ;  when  both  are  run- 
ning it  takes  18  hours  to  fill  it;  if  the  latter  is  closed  it  requires  15  hours  to 
fill  it :  if  the  former  is  closed,  how  long  will  it  take  the  latter  to  empty  it  7 

00.  Four  men,  A,  B.  C,  and  1),  spent  £'255,  and  agreed  that  A  should  pay 
| ;  B  £  ;  C  J  ;  and  D  }  :  how  much  must  each  pay  7 

07.  A,  B,  and  C,  formed  a  joint  stock  of  £820,  and  gained  £040.  in  the 
division  of  which  A  received  £5  as  often  as  B  did  £7,  and  C  £8:  how  much 
did  each  put  in  and  receive  7 

08.  A,  B,  and  C,  gained  a  certain  sum,  of  which  A  and  B  received  $040, 
B  and  C  $K80,  and  A  and  C  $800:  what  was  the  gain  of  each  7 

09.  What  number  is  that  -^  and  -J  of  which  being  multiplied  together,  will 
produce  the  number  itself? 

70.  A  club  spent  £2,  12s.  Id. ;  on  settling,  each  paid  as  many  pence  as  there 
were  individuals  in  the  party  :  how  many  were  there  in  the  party  7 

71.  The  sum  of  two  numbers  is  120,  and  the  difference  of  their  squares  ii 
4800  :  what  are  the  numbers  7 

72.  The  di (Terence  of  two  numbers  is  53.  and  the  difference  of  the  square* 
is  10759:  what  are  the  numbers  7 

73.  The  diagonal  of  a  square  is  80  ft. :  what  is  its  side  7 

•  74.  The  diagonal  of  a  square  field  is  120  rods:  what  is  its  are?  7 
75.  Find  the  side  of  the  greatest  square  beam  which  can  be  hewn..Srora 

log  5  ft.  in  diameter  7 

70.  The  mainmast  of  a  ship  is  95  ft.  long,  the  diameter  of  the  base  Js  3j  ft., 

that  of  the  top  2£  ft. :  what  is  its  solidity  7 

77    A  man  wished  to  tie  his  horse  by  a  rope  so  that  he  could  feed  on  just  an 

acre  of  ground  •  now  long  must  the  rope  be  7 


308  MISCELLANEOUS  EXAMPLES. 

78.  What  is  the  area  of  a  circle  1  mile  in  circumference "? 

79.  If  the  diameter  of  the  sun  is  887000  miles,  what  is  its  surface  1 

80.  If  the  diameter  of  Jupiter  is  86255  miles,  what  is  its  solidity  1 

81.  A  conical  stack  of  hay  is  20  ft.  high,  and  its  base  15  ft.  in  diametei . 
what  is  its  weight,  allowing  5  Ibs.  to  a  cubic  foot  1 

82.  How  many  bushels  will  a  cubical  bin  contain  whose  side  is  9  ft.  1 

83    How  many  hogsheads  will  a  cylindrical  cistern  10  ft.  deep  and  C^-   ft. 
Uameter  contain  1 

84.  How  far  from  the  end  of  a  stick  of  timber  30  ft.  long,  of  equal  size  from 
end  to  end,  must  a  lever  be  placed,  so  that  3  men,  2  at  the  lever,  and  I  at  the 
end  of  the  stick,  may  each  carry  i  of  its  weight  1 

85.  How  many  different  ways  may  a  class  of  26  scholars  be  arranged? 

86.  If  100  eggs  are  placed  in  a  straight  line  a  rod  apart,  how  many  miles 
must  a  person  travel  to  bring  them  one  by  one  to  a  basket  placed  a  rod  from 
me  first  egg  1 

87.  What  is  the  sum  of  the  series  1,  !•£,  2,  2|,  3,  &c.,  to  50  terms'? 

88.  A  blacksmith  agreed  to  shoe  a  horse  for  1  mill  for  the  first  nail  in  his 
shoe,  2  mills  for  the  second  nail,  and  so  on  :  the  shoes  contained  32  nails:  how 
much  did  he  receive  1 

89.  Said  a  mule  to  an  ass,  if  I  take  one  of  your  bags,  I  shall  have  twice  as 
many  as  you,  and  if  I  give  you  one  of  mine,  we  shall  have  an  equal  number: 
with  how  many  bags  was  each  loaded  1 

90.  What  number  taken  from  the  square  of  48  will  leave  16  times  541 

91.  Divide  $1000  between  A,  B,  and  C,  and  give  A  $120  more  than  C,  and 
C  $95  more  than  B. 

92.  A  person  being  asked  the  hour  of  the  day,  said,  that  the  time  past  noon 
was  f-  of  the  time  till  midnight :  what  was  the  hour  ] 

93.  A,  B,  and  C,  can  trench  a  meadow  in  12  days  ;   B,  C,  and  D,  in  14  days ; 
C,  D,  and  A,  in  15  days;  and  D,  B,  and  A,  in  18  days.     In  what  time  would 
it  be  done  by  all  of  them  together,  and  by  each  of  them  singly  1 

94.  Suppose  A,  B,  and  C,  to  start  from  the  same  point,  and  to  travel  in  the 
same  direction,  round  an  island  73  miles  in  compass,  A  at  the  rate  of  6,  B  of 
10,  and  C  of  16  miles  per  day :  in  what  time  will  they  be  next  together  1 

95.  At  what  time  between  12  and  I  o'clock  do  the  hour  and  minute  hands 
of  a  common  clock  or  watch  point  in  directions  exactly  'opposite  1 

96.  In  how  many  years  will  the  error  of  the  Julian  Calendar  involve  the 
loss  of  a  day  1 

97.  A  man's  desk  was  robbed  3  nights  in  succession ;  the  first  night  half  th« 
number  of  dollars  were  taken  and  half  a  dollar  metre  ;  the  second,  half  the  re- 
mainder was  taken  and  half  a  dollar  more;  the  third  night,  half  of  what  was 
then  left  and  half  a  dolfar  more,  when  he  found  he  had  $50  left,  how  much 
had  he  at  first  1 

THE    END. 


ANSWERS   TO  EXAMPLES. 


NOTE. — At  the  urgent  request  of  several  distinguished  Teachers,  who  hava 
received  Thomson's  Higher  Arithmetic  with  favor,  the  publishers  have  issued 
an  edition  of  it,  containing  the  answers  in  the  end  of  the  book.  It  is  hopea 
that  pupile,  who  may  use  this  edition,  will  have  suificient  regard  to  their  own 
improvement,  never  to  consult  the  answer  till  they  have  made  a  strenuous  and 
persevering  effort  to  solve  the  problem  themselves. 

N.  B. — The  work  without  the  answers  is  published  as  heretofoie. 

ADDITION.— ARTS.   59-61. 


Ex.      ANS. 

Ex.       ANS. 

Ex.      ANS. 

1.  $5445 

17.  288011295. 

35.  9429190. 

2.  41757  bushels. 

18.  14303433. 

36.  11178170 

3.  11  5  90  pounds. 

19.  100611775. 

37.  10306156. 

4.  831551. 

20.  1805851434. 

38.  10662291. 

5.  $5583. 

21.  337351. 

39,  40.  Given. 

6.  65440  sq.  miles. 

22.  7221.         J41.  214. 

7.  10-2451  sq.  m. 

23.  4251988. 

42.  253.   . 

8.  528524  sq.  m. 

24.  3795. 

43.  276. 

9.  650327  sq.  m. 

25.  73464390. 

44.  19443. 

10.  1362742  sq.  m 

26.  33604444. 

45.  20714. 

11.  233890. 

27.  15821984. 

46.  2476372. 

12.  828463. 

28.  97059404. 

47.  $132085946. 

13.  990240. 

29.  1038220930. 

48.  $107109740. 

U.  96181521. 

32.  $570805.       49.  2069857  tons. 

15.  127215713. 

33.  6460458  yards.  50.  $57981492. 

.6.  869754587. 

34.  6657039  pounds  51.  Given. 

SUBTRACTION.—  ART.  76. 

1.  $7095. 

10.  $12280043. 

19.  5313439. 

2.  28984  bushels. 

11.  $23563746. 

20.  543679. 

3.  $30954. 

12.  430143  tons. 

21.  2007984. 

4.  $46025. 

13.  149237. 

22.  45103074. 

5.  58000000  miles. 

14.  3393329. 

23.  66729549. 

6.  $6327597. 

15.  54399581. 

24.  72820280. 

7.  $26176670. 

16.  8825431. 

25.  55301760. 

8.  $1644737. 

17.  4001722. 

26.  80200180. 

9.  $7977899. 

18.  2601900. 

27.  95658143. 

400 


ANSWERS. 


[PAGES  46 — 60. 


SUBTRACTION    CONTINUED. A  AT.    76« 


Ex.       ANS. 

Ex.       ANS 

Ex.      ANS. 

28.  9000001. 

39.  85807625. 

50.  925. 

29.  99899999. 

40.  1598. 

51.  1511. 

30  83128433.      41.  4004. 

52.  41845. 

31.  40592424. 

42.  1384. 

)  $46900,  W. 

32.  55352005. 

43.  14061. 

5   )  $69450,  1! 

33,  19957466. 

44.  12494. 

54.  $2410  lost. 

81  77919201. 

45.  11547. 

55.  171825. 

35.  70051563.      46.  3295.         |56.  $1674737. 

30.  53201371.      j47.  1006. 

57.  $97. 

37.  25311703. 

48.  3707. 

58.  $3893. 

38.  86282745. 

49.  2004. 

59.  Given. 

MULTIPLICATION.—  ART.  93. 

1.  $24795. 

15.  3931476. 

29.  239968374861. 

2.  $36099. 

16.  415143630. 

30.  449148410434. 

3.  $56700. 

17.  31884470. 

31.  289975559744. 

4.  90520  miles. 

18.  8468670. 

32.  294144537440. 

.5.  74175  pounds. 

19.  43506216. 

33.  335834314400. 

6.  372500  days. 

20.  11847672.     J34.  18834782688. 

7.  960000  rods. 

21.  57380625.     \S5.  109588282050. 

8.  20835. 

22.  11050155200. 

36.  654638320927. 

9.  21576. 

23.  12810000. 

37.  396890151372. 

10.  68198. 

24.  48288058. 

38.  554270292192. 

11.  176400. 

25.  3473567604. 

39.  2985984. 

12.  1554768. 

26.  88789980848. 

40.  57111104051. 

13.  5497800. 

27.  9313702853. 

41.  60435595442394 

14.  1674918. 

28.  67226401140. 

42.  87112343040000 

CONTRACTIONS  IN  MULTIPLICATION.—  ARTS.  97—  1O8. 

8.  $1776. 

20.  312046700000.  '29.  96000  pounds. 

9.  $5760. 

21.  52690078000000  30.  359400000. 

10.  $8100. 

22.  6890634570000-  31.  143759940000. 

11.  5782  s. 

000. 

32.  28708635000)00 

12.  23808  miles. 

23.  494603050600- 

34.  123240000. 

13.  $11736. 

000000.       J35.  2309760000. 

14.  19845  s. 

24.  87831206507- 

36.  26366200000. 

15.  $32256. 

000000000. 

37.  144447000000. 

17.  46500  bushels. 

25.  678560051090- 

39.  31276000000. 

18.  365000  days. 

000000000. 

40.  3747600000000- 

19.  1534860000. 

28.  18750  pounds. 

41.  18054680000000 

PAGES  61 74.] 


ANSWERS. 


401 


PON  TRACTIONS    IN    MULTIPLICATION    CONTINUED. 


Ex.      ANS. 

Ex       ANS. 

Ex.      ANS. 

42.  664726500000- 

74.  4140. 

99.  180600000. 

000. 

75.  27936. 

100.  2722946304. 

43.  1075635900000- 

76.  154250. 

101.  2172069918. 

000. 

77.  11348400. 

102.  7225. 

4£  45514. 

78.  34639552. 

103.  65536. 

4F  68476. 

79.  2685942. 

104.  104650. 

47.  400624. 

80.  2801960. 

105.  127447'JO. 

48.  907002. 

81.  72156000. 

106.  31049291000. 

50.  132525. 

82.  1680000000. 

107.  2732116062240 

51.  307664. 

83.  2000000000. 

108.  222310980000 

52.  2333616. 

84.  43644865, 

109.  20066857745- 

53.  5691627. 

85.  81708550. 

896. 

55.  474309. 

86.  401939564. 

110.  1256700743298 

66.  6027966. 

87.  476413195. 

111.  37968807755. 

57.  7293699. 

88.  62220780. 

112.  39073U8478. 

58.  4629537. 

89.  637049231. 

113.  102128*493520 

63.  54530.   • 

90.  406101366. 

114.  1421400000000 

64.  72819. 

91.  42261696. 

115.  60302400000- 

65.  346896. 

92.  504159579. 

000. 

66.  6624403632. 

93.  6724232757. 

116.  9130020:1000- 

67.  17651712450.  < 

<94.  7306359. 

000. 

68.  21983532672. 

95.  21760506. 

117.  6800400^.0000- 

71.  625. 

96.  39429936. 

000. 

72.  2916. 

97.  2283344802. 

118.  4000000000- 

73.  5184. 

98.  650633256. 

000000 

DIVISION.— ART.   127. 


1.  45  bu.  13. 

2.  85  bbls.         14. 

3.  &68lfi-         15. 

4.  $3.  16. 

5.  I68ff.  17. 

6.  $73972|£i.  18. 

7.  20iff  days.  19. 

8.  m-iWds.    20. 

9.  2773||.         21. 

10.  1139V*.         22- 

11.  1443-s-V        .23. 

12.  1489ff.        124. 


5697f£. 

25.  3679. 

35.  826^51- 

3823-f-f. 

26.  4500. 

-iV^j'W' 

4166if. 

27.  50830TW- 

36.  1387805- 

21276|-f. 

28.  630. 

ww- 

12152^. 

29.  235. 

37.  900^00900- 

191-2*^. 

30.  648. 

90^9-j-^T. 

873. 

31.  267l01ff. 

38.  IJ009GOO- 

48. 

32.  563. 

oooo-KH. 

48HH- 

33.  8826211- 

39.  900H0900- 

87-iVs- 

Wf- 

009  rhr- 

108. 

34.  23434402- 

45- 

aV^"' 

402 


ANSWERS. 


[PAGES  75 — 96. 


CONTRACTIONS   IN   DIVISION.— ARTS.  129—139. 


fix.     AN^ 

Ex.     ANS. 

Ex     ANS 

Ex     ANS 

1,  2.  Given. 

18.  36. 

37.  15. 

55.  228378HI 

3.  132f|  a. 

19.  68. 

38.  16-ftV. 

56.  941501ff. 

4.  672. 

20.  36ff. 

39.  17. 

57.  478676if. 

5.  460. 

21.  75. 

40.  30. 

58.  59207. 

6.  205. 

23.  1207. 

41.  250  days. 

59.  1826896. 

7  1265. 

24.  1690. 

42.  950  years. 

60.  13S791-M 

8.  20;  34;  56d. 

25.  6512£. 

43.  $10-iV- 

61.  65964||i 

9.  1650;  7650; 

20.  8654. 

44.  $285f*8. 

62.  6162^5-. 

$43200. 

27.  83||. 

45.  $39rrV 

63.  1583  If-H. 

10.  267,  and 

28.  70if. 

46.  $2iff. 

64.  21i^ff.' 

50000  R. 

29.  77ff. 

47.  $ll-3Yr. 

65.  4134i|t. 

11.  144,  and 

30.  142if. 

48.  $54-3^. 

66.  3966|ff-. 

360791  R. 

31.  94. 

49.  $219ifi- 

67.  16581ff. 

12.  5823,  and 

32.  194|f. 

50.  18if. 

68.  7405-,^. 

67180309R 

33.  1693if. 

51.  13529||. 

69.  4362-iVs-. 

14.  105  b. 

34.  3795-2V 

52.  12466H- 

70.  3186/7V 

15.  184  bbls. 

35.  67. 

53.  12454|f.   '71.  9Vf$W. 

16.  197+f. 

36.  203-iVs- 

54.  13446913f  72.  920TfJHHhr 

CANCELATION.— ARTS.   15O,   151. 


2.  45. 

4.   65. 

6,  7.   Giten. 

9.   3i. 

3.   63. 

5.   73. 

8.  6. 

10.  3. 

APPLICATIONS  OP  THE  FUNDAMENTAL  RULES. 

ARTS.  152—159. 

1.  Given. 

11.  79.»years; 

19.  Given. 

27.  632. 

2.  255  acres. 

94  yrs. 

20.  48  beggars. 

28.  974. 

3.  925  bu. 

12.  $510|  car. 

21.  20  flocks. 

29.   7124;  5516 

4.  Given. 

$345i  hor. 

22.   Given. 

30.   13000; 

5.  $190. 

14.  65  years. 

23.   20  years. 

12264. 

6.   11  25  sheep. 

15.   175  rods. 

24.   10  months. 

31.   21151 

8.  $2240. 

17.   187825. 

25.    1842. 

20975. 

9.  $3436. 

18.   1033062. 

26.   1062. 

32.   786. 

PROPERTIES  OF   NUMBERS.—  ARTS.   162,   163. 

1—9.  Given. 

13.   2024122. 

18.   707961. 

22.   1614386. 

10,  20212331. 

14.   1522365. 

19.   1036993. 

23.    118620366. 

11.   2350147. 

15.  Given. 

20.  9753020. 

24.   3879090- 

12.   1331124. 

16,   17.  Given.  21.  360913096. 

582. 

?AGES  97,  98.J 


ANSWERS. 


403 


ANALYSIS   OP  COMPOSITE  NUMBERS.— ART.   165. 


Ex.      ANS. 

Ex.       ANS. 

Ex.      ANS. 

4.  9  —  3X3. 

28.  2,  3,  and  7. 

52.  2,  and  37. 

5.  2,  and  5. 

29.  2,  2,  and  11. 

53.  3,  5,  and  5. 

6.  2,  2,  and  3. 

30.  3,  3,  and  5. 

54.  2,  2,  and  19. 

7.  2,  and  7. 

31.  2,  and  23. 

55.  7,  and  11. 

8.  3,  and  5. 

32.  2,  2,  2,  2,  and  3. 

56.  2,  3,  and  13. 

9.  2,  2,  2,  and  2. 

33.  7,  and  7. 

57.  2,  2,  2,  2,  and  6 

10.  2,  3,  and  3. 

34.  2,  5,  and  5. 

58.  3,  3,  3,  and  3. 

11  2,  2,  and  5. 

35.  3,  and  17. 

59.  2,  and  41. 

12  3,  and  7. 

36.  2,  2,  and  13. 

60.  2,  2,  3,  and  7. 

13.  2,  and  11. 

37.  2,  3,  3,  and  3. 

61.  5,  and  17. 

14.  2,  2,  2,  and  3. 

38.  5,  and  11. 

62.  2,  and  43. 

15.  5,  and  5. 

39.  2,  2,  2,  and  7. 

63.  3,  and  29. 

16.  2,  and  13. 

40.  3,  and  19. 

64.  2,  2,  2,  and  11. 

17.  3,  3,  and  3. 

41.  2,  and  29. 

65.  2,  3,  3,  and  5. 

18.  2,  2,  and  7. 

42.  2,  2,  3,  and  5. 

66.  7,  and  13. 

19.  2,  3,  and  5. 

43.  2,  and  31. 

67.  2,  2,  and  23. 

20.  2,  2,  2,  2,  and  2. 

44.  3,  3,  and  7. 

68.  3,  and  31. 

21.  3,  and  11. 

45.  2,  2,  2,  2,  2,  and  2 

69.  2,  and  47. 

22.  2,  and  17. 

46.  5,  and  13. 

70.  5,  and  19. 

23.  5,  and  7. 

47.  2,  3,  and  11. 

71.  2,  2,  2,  2,  2,  and  3 

24.  2,  2,  3,  and  3. 

48.  2,  2,  and  17. 

72.  2,  7,  and  7. 

25.  2,  and  19. 

49.  3,  and  23. 

73.  3,  3,  and  11. 

26.  3,  and  13. 

50.  2,  5,  and  7. 

74.  2,  2,  5  and  5. 

27.  2,  2,  2,  and  5. 

51.  2,  2,  2,  3,  and  3. 

75.  2,  2,  3,  3,  and  3. 

76.  120=2X2X2X3X5. 
144=2X2X2X2X3X3 

77.  180  =  2X2X3X3X5. 
420=2X2X3X5X7. 

78.  714=2X3X7X17. 
836  =  2X2X11X19. 

79        574  =  2X7X41. 

2898  =  2X3X3X7X23. 

80.  11492  =  2X2X13X13X17 

980=2X2X5X7X7. 

81.  650  =  2X5X5X13. 
1728  =  2X2X2X2X2X2 

X3X3X3. 


82.     1492  =  2X2X373. 

83. 
84. 

85. 
86. 


251. 

4604=2X2X1151. 
16806  =  2X3X2801. 
71640  =  2X2X2X3X3X5 

X199. 

20780=2X2X5X1039. 
84570  =  2X3X5X2819. 
65480  =  2X2X2X5X1637 
92352  =  2X2X2X2X2X2 

X3X13X37. 
81660=2X2X3X5X1361, 


T.H. 


18 


404 


ANSWERS. 


[PAGES  99 — 118 


GREATEST  COMMON   DIVISOR.— ARTS.  168—171. 


Ex.           ANS. 

Ex.           ANS. 

Ex.            ANS. 

Ex.            AN». 

1.  Given. 
2.  3. 
3.  7. 
4.  5. 
6.  2. 

6,  7.  Given. 
8.  15. 
9.  14. 
10.  111. 
11.  39. 

12.   1. 
13.  Given. 
14.  3. 
15.   16. 
17.   15. 

18.    12. 

19.  18. 
20.  35. 
21.  6. 
22.  28. 

LEAST  COMMON  MULTIPLE.— ARTS.  176,  177. 


1—3.  Given. 

4.  90. 

5.  144. 

6.  180. 

7.  360 


8.  720. 

9.  12600. 

10.  504. 

11.  1134. 

12.  15015. 


14.  144. 

15.  600. 

16.  2520. 

17.  252. 

18.  1134. 


19.  360. 

21.  600. 

22.  1440. 

23.  13824. 

24.  51000. 


REDUCTION   OF   FRACTIONS. —ARTS.  195-2O1 


1,  2.  Given. 

3.  i. 

4.  f 

5.  -£-. 

6.  f. 

7.  f 

8.  |f. 

9.  f*. 

10.  T^. 

11.  ii- 

12.  i. 

13.  li 

14.  i. 


15.  |. 

16.  f. 
17. 
18. 

21.  9. 

22.  5. 

23.  3f. 

24.  9|. 

25.  1. 

26.  60. 

27.  21. 

28.  52. 

29.  60f. 


W;  * 
ifil-f; 


63. 
64. 
65. 
66. 
67. 
68. 
69. 
70. 
VI. 

72.  -mm; 

73,  74,  Given. 


ifH;  iffX; 
;  fm* ;  ii 
:  HffS; 


30. 

34. 
35. 
36. 
37. 
38. 
39. 
40. 
41. 
42. 
43. 
44. 

75. 

76. 
77. 
78. 
79. 
80. 
81. 
82. 
83. 
84. 
85. 


23  6  g6 

ni*.' 

J5JLJJL 


47. 
48. 
49. 
52. 
53. 
54. 
55. 
56. 
57. 
58. 

59.  f 

60.  -ft. 

61.  62    Given 


il; 

i£; 

JL&  • 
If  4    > 

if; 


*;  If; 

a  •    J2-L 
4   >     2  4« 


-ii;  -A-; 
H;  tt; 

0  U    J      CU>      GU>      60* 

ft;  ft;  n;  ft- 
t&**;  ^;  f"' 

1  o  o  s  >   TSnJe  >    Too*  t    i  oo  f» 


PAGES   118—127.] 


ANSWERS. 


405 


REDUCTION    OF    FRACTIONS    CONTINUED. ART.  201. 


El. 

80. 

87. 
88. 


ANS. 


ttftt;  ttftt;  nm; 
fHft- 

-23d.tl«    -2U.QJL'    _J>  8  Q_  •    ^.lAO. 
C  3  II  0  >     63  00   >      63UO»      6  3  U  0  • 


Ex. 


ANS. 


89. 
90. 


ft;  -Aft;  £ft;  •¥& 

;  ttf  i  tH;  ill 


ADDITION  OF   FRACTIONS.—  ARTS.  202-2O  i. 


1  —8.  Given. 
4    3^a 
5 .   2  2  T. 
0. 


2  ^. 

mt. 


9. 
10. 
11. 
12. 


23.  •£&  ; 

24.  -f^; 

25.  iif ; 
20.   14-f; 
27.   IvVV 

30.  Ml* 
3 1  ^  A  &  a  a 


5.   gV- 

0.   5V=i. 

7.  MV- 

8.  sWi- 

9.  i,VA. 


13.  1-fr. 

14.  /4V 

15.  2^. 
10.  214* 

17.  114-f 

18.  G|. 

19.  01  f. 

20.  15^ 

017  _6.A_    nl 


SUBTRACTION  OF   FRACTIONS.— AUTS.  2O6  208. 

10.  iiJ.  17.  121f£.    25.  291 

11.  ii.  18.  278|i     2«.  0034^. 

12.  23.  21. 

13.  -LSA-  oo 

15. 


10. 


23.  137f. 

24.  4GG* 


28.  -f. 

29.  8203 a. 


30.   7  H|. 


MULTIPLICATION  OF  FRACTIONS.—  CASE  I.—  ARTS.  21  1     IT. 


1—3.  Given. 

17.  G33|. 

32.  514^. 

48.  0897. 

4.  ^=3^. 

18.  3715*. 

33.  3  05  |f  i 

49.  15282. 

5.  ^s  =  12i-. 

19.  444  Sf. 

35.  10390!-. 

50.  29318. 

D  .  18  \  i  . 

20.  22G4,V 

30.  174GOH. 

51.  1280. 

7.  37?. 

21.  12519*. 

37.  SCGVaV 

52.  279. 

8.  48H- 

24.  108. 

38.  20G73fy. 

53.  4490. 

9.  80-3V. 

25.  127. 

39'.  35650|-f 

54.  8113. 

10.  GGf?f. 

20.  240. 

40.  23555  HI. 

55.  10413-!*,. 

11.  1  G  i  T  i  "$"• 

27.  435. 

43.  375. 

50.  5072  ff. 

12.  770-6V 

28.  500. 

44.  738. 

57.  5080!*,. 

13.  G02ifi. 

29.  02  H. 

45.  1178. 

58.  43452. 

14,  15.  Given. 

30.  701|. 

40.  3450. 

59.  74290iV 

16.  735?. 

31.  76A- 

47.  0795. 

CO.  92280ff. 

1 


406 


ANSWERS.  [PAGES  .  28 — 142. 


MUOIPLICATION    OF    FRACTIONS    CONTINUED. CASE    II. 

ARTS.  219,  22O. 


ill.           ANS. 

Ex.           ANS. 

Ex.            ANS. 

Ex.            ANS. 

1,  2.  Given. 

11.    17. 

20.  3232f&. 

29.  99|. 

3.  If  =  tV- 

12.  53|. 

21.  5143-iWT- 

30.  78f. 

±    ^&==^fet 

13.  242ff. 

22.  6998^-|4l^. 

31.  47tV. 

5.  -iVr- 

14.  329. 

23.   167if|. 

32.  86. 

6.  i. 

15.   1362f. 

24.  53TVs. 

33.  401f. 

>jf  ^f£f. 

16.  3198if. 

25.  24091|. 

34.  1474^V|, 

8.LJL& 
.    "|  7  5  . 

17.  451^i. 

26.  466-M-. 

35.  $972f. 

9.  ff  . 

18.  834|^f. 

27.  300000. 

36.  $6323^. 

10.  Given. 

19.  20011ai1r. 

28.  700. 

37.  5501-!^  na. 

CONTRACTIONS  IN  MULTIPLICATION  OP   FRACTIONS. 
ARTS.  221-225. 


3.  i.         . 

11.  f. 

21.  57600. 

32    4762^-. 

4.  A. 

12.  -4V 

22.  99000. 

33    J5937£ 

6.  i. 

13.  i. 

23.  187l|. 

34    40187^ 

6.  -¥-=2f. 

1'4.  A- 

24.   14220. 

35    65450. 

*  -  f. 

15.  A. 

26.  2133^. 

37     1278i. 

8.  A- 

16.  *VV 

27.  18466|. 

38    4083^. 

9.  i. 

19.  493£. 

28.  5580. 

39.  8150. 

10.  26. 

20.  8533^ 

29.  430000. 

40.  93833|. 

DIVISION  OP  FRACTIONS.—  ARTS.  226—241. 

1—3.  Given. 

18,  19.  Given. 

33.  9-tV*. 

51.  A- 

4.  A- 

20.  H. 

34.   13iif. 

52.  f. 

5-  *¥*• 

21.  A- 

37.  13-tWr. 

53.  3H. 

64              g 
•     TO^^S' 

22.  23|. 

38.  STWir- 

54.  fr. 

7.  A- 

23.  40£. 

40.  220^. 

55.   1?. 

8.  A=*. 

24.  i. 

41.  30fi. 

56.   1. 

o.  vw- 

25.  i. 

42.  50H- 

58.   Hf. 

10—13.  Given. 

26.  3ff. 

43.  6^. 

59.  A- 

14.  1AV- 

29.   135. 

46.  383-ftV- 

60.  f;. 

15.  2if. 

30.  168. 

47.  54VVVV. 

61.  5|. 

10.  3H. 

31.   11A- 

49.  2fj. 

63.  ^ 

17.  m. 

32.   llf|. 

50.  -H. 

64.  i 

APPLICATION  OF   FRACTIONS.—  ART.  242. 

1.  88if  yds. 

4.  $14if. 

7.  $1548|. 

10.  5606^  s. 

2.  162tVlbs. 

5.  $62H. 

8.  $55  16£. 

11.  $483f. 

3.  $54f. 

0.  $635^-. 

9.  $1515". 

12.  $100. 

PAGES   143 170.] 


ANSWERS. 


407' 


APPLICATION    OF    FRACTIONS    CONTINUED. ART.    242. 


Ex             ANS 

Ex.             ANP. 

K.\              ANS 

Ex              AN* 

13.  &16111-J-. 
14.   404552fflb 
15.   -Si  8061. 
16.  36121  bu. 
17.  $30968$-. 
18.  G939-rVm. 
19.  5229  m. 
20.  $9l7oi. 

21.   lOoi  yds. 
22.   218|  Ibs. 
23.  626|  gals. 
24.   20-rVlbs. 
25.  53-f  yds. 
26.  27  boxes. 
27.   153f  bbls. 
28.  2911  suits. 

29.  3  8  A  rods. 
30.  $3f-g£. 
31.   $6f. 
32.   584f  bu. 
33.  4|  doz. 
34.   6-6\  cts. 
35.   9|fi  s. 
36.  $13-,V*. 

37.  &5  Hi- 
38.  42-,-fr  tons. 
39.  SlYVVr. 
40.  t51+i«*. 
41.  $3fg£. 
42.  2664-Hr  d 
43.  $33798J|f. 

REDUCTION.— ART.  282. 


2.  68810  far. 

28.  8553600  in. 

52.-  28992  pts. 

3.  86768  far. 

29.  5280000  yds. 

53.  1427  bu.  1  pk. 

4.  284079  far. 

30.  54  m.  7  fur.  38  r. 

54.    130  100  qts. 

5.  966  15  far. 

2  yds.  2  ft. 

55.  36360  inin. 

6.  £'25,  13s.  6d.  3  far. 

31.  91.2m.  4  fur.  31  r. 

56.  3  1557600  sec. 

7.  £433,  Is.  2d.  3  far. 

1  £  yds.  2  ft,  7  in. 

57.  84  wks.6hrs.  45  mm 

8.  '266  guin.  18s.  8d. 

32.  5031  rods. 

58.  65  d.  2h.  4m.  40  sec 

9.   1448  sixpences. 

33.    17  m.  20  r. 

59.  3  1556928  sec. 

10.  6050  threepences. 

34.  132105600ft. 

60.  946728000  sec. 

11.   170472  grs. 

35.  2560  na. 

61.   10  yrs. 

12.  9000  pwts. 

36.  5000  qrs. 

62.  397200". 

13.   1010047  grs. 

37.  6396  yds.  2  qrs.  Ina  63.   1350000." 

14.  2  Ibs.  1  oz.  10  pwts. 

38.  9302F.e.4qrs.3na 

64.  2126°,  IT,  54". 

16  grs. 

39.  10156  na. 

65.  555555s.  16o,40'. 

15.  1  77  Ibs.  9  oz.  12  pwts 

40.  7116  qts. 

66.  470660  sq.ft. 

16.   1596  Ibs. 

41.  693  gals. 

67.  43660734  sq.  ft. 

17.  564000  oz. 

42.  26528  gi. 

68.  32640858360  sq.  in. 

18.   104300  Ibs. 

43.  48  bar.  20  gals. 

69.  582  A.  1  R.  3  r. 

19.  71  680000  drs. 

44.  117  pi.  1  hhd.  46  g. 

269}  sq.  ft. 

20.  lOcwt.  16  Ibs. 

3  qts.  1  pt.  2  gi. 

70.  259200  cu.  in. 

21.  133T.  12cwt.351bs 

45.   102  128  gi. 

71.  4551  552  cu.  in. 

22.   1  T.  202  Ibs.  1  oz. 

46.   12960  pts. 

72.  10877760  cu.  in. 

23.  9  120  drs. 

47.  87  bar.  26  gals. 

73.  49  cu.  ft.  1  cu.  in. 

24.  37440  sc. 

48.  630  hhds.  44  gals 

74.  306  C.  48  cu.  ft. 

25.  64  Ibs.  11  oz.  5  drs. 

49.  19520  pts. 

75    4492800  cu.  in. 

26.  881bs.4oz.7drs.2sc 

50.  488  qts. 

76.  52  T.  40  cu.  ft. 

27.  142560ft. 

51.  24440  qts. 

180  cu.  in. 

APPLICATIONS   OF   REDUCTION.—  ARTS.  283—  2§i. 

1.  Given. 

4.  177-flbs.  Troy,  or 

7.  5  8  Ibs.  4  oz.  Troy. 

2.  576  Ibs.  avoir. 

145-^f-f  Ibs.  avoir. 

8.  21f^2-  Ibs.  Troy. 

3.  691  Ibs.   10  oz. 

5.   265f  Ibs.  Troy,  or 

9.  271  Ibs.  3  oz. 

5-rVr  drams. 

218ifilbs.avoir. 

10.  Given. 

408 


ANSWERS.  [PAGES  171- -180 


APPLICATIONS  OF  REDUCTION  CONTINUED. ARTS.  285—293. 


F.«.              Ass. 

E.x.               AN*. 

K.\.                ANS. 

!1.    3  00  sq.ft. 

30.   3450  wine  gals. 

49.   14  hhds.  4SK  g* 

12.    14  A.  10  *q.  rds.|3l.   8040f  w.  gals. 

51.  51^f  beer  gals. 

13.    lOSsq.  v.  8  sq.  ft. 

32.  5184  beer  gals. 

52.  45  'ji  wine  gals. 

14.  440  A.  1  It. 

33.  09  1  2  b.  gals.  2  qts 

53.   24f  ?  w.  gals. 

15.  40  A. 

34.   80  i6}-  l>u. 

54,   1598  w.  gals. 

10    30  sq.  yds. 

35.    100  bu. 

55.   2207|-f  w.  gals. 

17    60  sq.  yds. 

36.  800  bu. 

56.   3125H  qts. 

13.    lll|sq.  yds. 

37.  897-H-  w.  gals. 

57.  2734*?  gals. 

20.   56  1  cu.  ft. 

38.   7l2fff  bar. 

59.  8  min.  36  sec. 

21.   120  cu.  ft. 

39.  902807H1  hhds. 

60.  39  min. 

22.   80  0.  2  cu.  ft. 

41.  622f  cu.  ft. 

61.   1  hr.  8  m.  40  sec. 

23.   748  cu.  ft.             142.    1244f  cu.  ft. 

62.  33  min.  48  sec. 

24.  750  cu.  ft. 

43.  8*1  cu.  ft. 

63.   12h.28m.  12s. 

25.   72  cu.  yds. 

44.   210^  cu.  ft. 

64.  Given. 

26.   160  cu.  ft. 

45.  842-ft  cu.  ft. 

65.  4°  45'. 

27.   1800  cu.  ft. 

47.  42-/A  bu. 

66.   12°  46'. 

29.  17280  bu.             !48.  40-ft-  gals. 

67.   13°  23'. 

COMPOUND   NUMBERS  REDUCED  TO  FRACTIONS.— ART.  290. 


1-4.   Given. 
5. 

0. 


8.  -ft  lb.  Troy. 

9.  eWlb.Troy 
10.  if  lb.  avoir. 

HJidi  T 
.     4  0  U     *• 


1-2.   -f  yd. 
13.  -ffjm. 
14. 
15 


16. 
17. 
18. 
19. 


A. 
r/r  sq.  r. 


igal. 
i  hhd. 

d. 

hr. 


20.  TTTsYinr. 
22.  VV- 
23    -£-0.- 

24!  I' 

25.  -gV 

26.  fy. 

27.  /0\. 
28. 


29. 
30. 
31. 
32. 
33. 
34. 
35. 
36. 


FRACTIONAL    COMPOUND    NUMBERS 
'REDUCED  TO  WHOLE  NUMBERS  OF  LOWER  DENOMINATIONS. — ARTS.  297,  298t 


3. 

17s.  Gd. 

13. 

2  qts.  1  pt 

•Hgi- 

25. 

6£f  hrs. 

4. 

7d.  i  far. 

14. 

55  gals.  1 

pt. 

26. 

2688  min. 

5. 

5  oz.  2  p.  2 

°7    g- 

16. 

3  pks.  1  qt. 

Hpts. 

27. 

8-6\  na. 

G, 

12  pwts.  1 

2  grs. 

17. 

46  min.  40 

sec. 

28. 

17^-?-  qts. 

7. 

10  oz.  10| 

drs. 

18. 

21  hrs.  36 

min. 

29. 

174H?  qts, 

S 

57  Ibs.  2oz. 

4|drs. 

19. 

22£  sec. 

30. 

4|?  oz. 

9 

1250lbs. 

20. 

17'  Sf. 

31. 

60  pwts. 

10 

2  ft.  4f  in. 

22. 

Wd. 

32. 

7-j^-  r. 

11. 

6  ft.  2|  in. 

23. 

-aVy-  oz. 

33. 

A  sq.  ft. 

12. 

177  r.  12  ft  10  in. 

24. 

Hr. 

34. 

70'' 

PAGES   182     -190.  J  ANSWERS. 


409 


COMPOUND   ADDITION.— ART.  3OO. 


Ex.       ANS 

Ex.       ANS 

Ex.      ANS. 

3.  £106,  3s.  Id. 
4.  £188,  13s.  id. 
5.  9  T.  8  cwt.  17  Ibs. 
6  45T.  4  cwt.  57  Ibs. 
2  oz. 
7  1  07  Ibs.  7  0.8.  p.  Ig. 
8  330  Ibs.  2  o.  3  p.  5  g. 
9.  4  fur.  1  3  r.  13  ft.  Sin. 

10.  1091.  2m.  6  fur.  1ft. 
11.  114  yds.  3  qrs. 
1  2.  387  yds.  1  qr. 
13.  138  A.  114  sq.  r. 
80  sq.  ft. 
14.  468  A.  1R.  33sq.r. 
15.  43  sq.  yds.  5  sq.  ft. 
125  sq.  in. 

16.  240  gals. 
17.  181  hhds.  59  gals 
1  pt.  1  gi. 
18.  115  w.  15  h.  25m. 
19.  322  bu.  1  pk.  5  qts. 
20.  135  qrs.  3  bu.  3  pka 
2  qts. 

COMPOUND   SUBTRACTION.— ARTS.  3O2,  303. 


1.  Given. 

2.  £9.  2s.  8d.  3  qrs. 

3.  £60,  4s.  7d.  3  qrs. 

4.  £499, 1 3s.  4d.  2  qrs. 

5.  8  cwt.  1  qr.  6  Ibs. 
10  oz. 

6.  24  T.  1  cwt.  71  !ba. 

7.  19m.  289   .  2ft. 

8.  1  1.  1  m.  7  fur.  lOr. 


ft. 


9.  35  bu.  2  pks.  6  qts. 

10.  19  qrs.  6  bu.  2  pks. 

11.  55  yds.  2  qrs.  3  na. 

12.  44yds.  1  qr.  3  na. 

13.  6  gals.  2  qts.  1  pt. 

14.  48  hhd.  46  g.  2  qts. 

15.  85  A.  119r. 

16.  235  A.  48  r. 

17.  56  C.  90  cu.  ft. 

18.  339  cu.  ft.  26  in. 


19.  25°  3'  15". 

20.  35°  3'  30" 

21.  10°  26'. 

22.  54  yrs.  2  mos.  2  wka 
6d.2hrs.45min.6s 

23.  Given. 

24.  67  yrs.  9  mos.  22  d 

26.  1  yr.  5  mos.  lid. 

27.  3  yrs.  9  mos.  22  d. 


COMPOUND   MULTITPLICATION.— ART.  305. 


1,  2.  Giv*en. 

3.  £247,  6s.  Id. 

4.  £24,  9d. 

5.  17T.  55  Ibs. 

6.  403  T.  17cwf.551bs 

7.  689  Ibs.  8  oz.  16pwts 

8.  6  Ibs.  lOoz.  lOpwts 

9.  3039  hhds.  39  gals. 
1  f|t.  1  pt. 

10.  5668  pi.  32  gals. 

11.  2358  yds. 

12.  5375  yds. 

13.  14778  m.  1  fur.  32  r. 


14. 

15. 

ifi. 

17. 
18. 
19. 

20. 
21. 

22. 


2044  1.  1  m.  4  fur.  23. 
30  r.  |24. 

8962  bu.  16  qts.  J25. 
2968  qrs.  5  bu.  2  pks.  26. 
6  qts.  27. 

7821  A.  20  r.  28. 
25172  A.  1  R.  3r  29. 
24645  cu.  ft.  930  30. 
en.  in.  31. 
96350  C.  50  cu.  ft.  ' 


12783d.  lib.  28m. 
1199  yrs.  9  mos. 
3  wks.  Id. 


15891°  13'  30" 
204°  10'. 
4581  bu.  8  qts. 
2453  bu.  4  qts. 
£5,  16s.  10  id. 
£679,  3s.  4d. 
£297. 

£507,  16s.  3d. 
36  C.  74  cu.  ft. 
944  in. 

865  Ibs.  12  oz. 
25418  Ibs.  12  07. 
8662  gals.  2  qts. 


COMPOUND   DIVISION.— ART.  307. 

8.  £4,  17s.  3d.  i  qr. 


1-3.  Given. 

4.  51  Ibs.  3  oz.   10  pwts. 
15-f  grs. 

5.  31  bu.  14|-  qts. 

6.  25  bu.  I1  pts. 
7    £20.  Is.  6d. 


9.   10  yds.  3  qrs.  If  na. 

10.  9  yds.  1  qr.  i|  na. 

11.  83  m.  2  fur.  26  r.  11  ftv 

12.  214  m.  2  fur.  27  r. 
4  ft.  4  in. 


410 


ANSWERS.  [PAGES  196 — 201 


COMPOUND    DIVISION    CONTINUED. ART.    307. 


Ex.                            ANS. 

Ex.                            ANS 

13.   1  gal.  2  qts.  1  pt.  lH  gi. 

18.   Is.  17°  52'  21|f". 

14.   44  hhds.  29  gals.  1  pt.  l£gi. 

19.   90.  84  ft.  101  6W  in. 

15.  24  d.  8  hrs.  42  min.  40  sec. 

20.  6  C.  92  ft.  850|i  in. 

16;   10  yrs.  35  d.   1  hr.   13  min. 

21.  6s.  10£d. 

llT*j-sec. 

22.  7s.  lid.  3^qrs. 

17.   1°  48'  41i". 

23.   10s.  lid.  2-74qrs. 

ADDITION  OP   DECIMALS.—  ART    32O. 

],  2.  Given. 

9.  857.005.                    |16.  2.471092. 

3.  4-28.1739. 

10.   1097.84143. 

17.  0.0711824. 

4.  103.85-23. 

11.  1408.25559. 

18.  0.3532637. 

5.   14.747274. 

12.  127.05034. 

19.  0.807711. 

6.  60.149. 

13.  33.3182746. 

20.  0.1627165. 

7.  332.1249. 

14.  15674.1613. 

21.  0.996052. 

8.  501.15998. 

15.  1.807. 

22.  0.329773. 

SUBTRACTION  OF   DECIMALS.—  ART.  3J2. 

1,  2.  Given. 

13.  2.291. 

24.  0.000999. 

3.  1427.633782. 

14.  9.9999999. 

25.  699.93. 

4.  20.987651. 

15.  8.000001. 

26.  28999.908. 

5.  72.5193401. 

16.  4635.5346. 

27.  255999909.744. 

6.  81.16877. 

17.  541.787. 

28.  0.414. 

7.  0.066721522. 

18.  46.43606. 

29.  0.0041. 

8.  0.01. 

19.  0.0000999. 

30.  0.000000000999. 

9.  9.999999. 

20.  0.0000396. 

31.  0.002873789. 

10.  64.0317753. 

21.  31.99968. 

32.  0.062156. 

11.  24680.12377. 

22.  44.99955. 

33.  0.71699. 

12.  24.75. 

23.  98.99999901. 

34.  0.0000843174. 

MULTIPLICATION   OF   DECIMALS  —ART.  324. 

1.  681.45ft. 

13.  36.740232. 

25.  0.00164389993. 

2.  25020  miles. 

14.  919.82036. 

26.  160.86701632806. 

3.  2055.375  gals. 

15.  0.000000072. 

27.  0.06288405909156. 

4.   136.125  nails. 

16.  0.00105175. 

28.  2.5067823. 

5    788.0125  sq.  yds. 

17.  390.657556. 

29.  64.327106105314. 

6    43560  sq.  ft. 

18.  275.230594. 

30.  0.0000118260069. 

7.  2465.375  sq.  rods. 

19.   148.64244532. 

31.   11027.10199543710 

8.  0.250325, 

20.  73.25771882. 

32.  94167471.869654- 

9.   18.93978. 

21.  52.17977576. 

039. 

10.   14.78091. 

22.  0.0306002448. 

33.  .OOOOf'006676542- 

11.  0.613836. 

23.  4701.169144360. 

672. 

12.  0.0320016. 

24.  536.660(175952. 

PAGES  201 213.]  ANSWERS. 


411 


CONTRACTIONS  IN  MULTIPLICATION  OF   DECIMALS. 
ARTS.  325-32T. 


Ex.                ANS. 

Ex.               ANS. 

Ex.                ANS. 

1.  Given. 
2.  4:29302.13401. 
3.  1067123.50123. 
4.  6)8340.17. 
5    304672.14067. 
6.   44632140.32. 
7.  2134567.82106. 
8    500. 

9.  75000. 
10.  6.5. 
11.  48. 
12.  2480. 
13.  381. 
14,  65.04. 
15.  834000. 
16.  10. 

17-20.  Given. 
21.  0.09484. 
22.   1.262643. 
23.  0.0769. 
24.  0.0389254. 
25.  0.00876. 
26.  0.002516. 
27.  0.001789. 

DIVISION   OF   DECIMALS.    ART.  33O. 


1-3.  Given. 

4.  13  boxes. 

5.  8  suits. 

6.  4.98347+days. 

7.  82.9997-Hoads. 

8.  27.7173-Hays. 

9.  150.25  bales. 

10.  5.929 1-f. 

11.  6.632. 


12.  79098.8235-f 

13.  0.6344-h 

14.  1210.2344+. 

15.  0.03. 

16.  1 34.8805+. 

17.  59.4060+.  . 

18.  24.8266-f. 

19.  4320.67. 

20.  0.02. 


21.  83671000. 

22.  255.1210-J-. 

23.  0.000005. 

24.  60.2589. 

25.  211.076. 

26.  400000. 

27.  60000000. 

28.  4000000. 

29.  3 11. 487360+. 


CONTRACTIONS  IN  DIVISION  OF  DECIMALS.— ARTS.  331-33 


1,  2.  Given. 

3.  67234.567. 

4.  103.42306. 

5.  0.42643621. 

6.  6.72300045. 


7.  0.000012300456. 

8.  0.0000020076346. 

9.  Given. 

10.  0.1274. 

11.  0.09471. 


12.  1.611. 

13.  0.04026. 

14.  0.0954776. 

15.  2.0208. 

16.  0.980439. 


DECIMALS  REDUCED  TO  COMMON  FRACTIONS.— ART.  335. 


1,  2.  Given. 

3.  i. 

4.  if 

5.  V. 


10. 
11. 
12. 
13. 


'14. 
15. 
16. 


COMMON   FRACTIONS    REDUCED   TO    DECIMALS. 
ARTS.  337—344. 


A-3.  Given. 

10. 

0.6. 

4.  0.5. 

11. 

0.8. 

5.  025. 

12. 

0.5. 

6.  0.5. 

13. 

0.125. 

7.  075. 

14. 

0.25. 

8.  02. 

15. 

0.375. 

9.  0.4. 

16. 

0.5. 

18* 

17.  0.625. 
18.  0.75. 
19.  0.875. 
20,  21.  Given. 

25.  Terminate 
26.  Terminate. 
27.  Interminate, 
28.  Terminate. 

22.  Terminate. 

31.  03. 

23.  Terminate. 

32.  0.6. 

24.  [nterminato. 

33.  0.16. 

ANSWERS 


[PAGES  214  —  224 


COMMON    FRACTIONS    REDUCED    TO    DECIMALS    CONTINUED. 


Ex.     ANS. 

Ex.     ANS. 

Ex.     ANS. 

Ex.     ANS. 

34.  0.3. 

41.  0.714285. 

48.  0.6. 

56.  0.0048828- 

35.  0.6. 

42.  0.857142. 

49.  0.7. 

125. 

36.  0.83. 

43.  0.1. 

50.  0.8. 

57.  0.583. 

37.  0.142857. 

44.  0.2. 

51.  0.1875. 

58.  0.076923. 

38.  0.285714. 

45.  0.3. 

52.  0.076923. 

59.  0.0104895. 

39.  0.428571. 

46.  0.4. 

53.  0.024. 

60.  0.46835443- 

40  0.571428. 

47.  0.5. 

54.  0.0112. 
55.  0.275. 

03797. 

COMPOUND  NUMBERS  REDUCED  TO  DECIMALS.— ART.  3  16. 


1,  2.  Given. 

3.  £0.5375. 

4.  £0.825. 
6.  £0.87916. 


6.  0.416  s. 

7.  0.5416  s. 

8.  0.115625  m. 

9.  0.25625m. 


10.  0.2583  hr. 

11.  0.127083  d. 

12.  0.0525  cwt. 

13.  0.46875  Ib. 


14.  0.875  bu. 

15.  0.5625  pk. 

16.  1.125  gals. 


DECIMAL   COMPOUND   NUMBERS    REDUCED   TO    WHOLE 
ONES.— ART.  348. 


2.  14s.  6d. 

3.  2s.  7d.  3.2  qrs. 

4.  Id.  2  qrs. 

5.  9d.  3.6  qrs. 

6.  12  Ibs.  8  oz. 


7.  6  oz.  15.36  drs. 

8.  88  rods. 

9.  7  ft  0.51  in. 

10.  11  gals.   1  qt.  1  pt. 
3.7184  gills. 


11.  1  qt.  Ipt.  3.4432  gi, 

12.  10  h.  13m.  9. 12  sec. 

13.  50  min.  42  sec. 


REDUCTION  OP  CIRCULATING  DECIMALS.— ARTS.  355—61. 


1,  2.  Given. 
3.  if,  or  -ft. 
4. 

5. 


9. 
10. 
14. 
15. 
16. 


-9V9,or-rh;.l7. 
18. 
19. 


20.  T| 
or 
23.  0.277. 


0.333. 
0.045. 
24.  4.3213. 
6.4263. 
0.0000. 


7.  Vkor-ji 

ADDITION  OF  CIRCULATING  DECIMALS.— ART.  362. 


2.  179.2745563. 

3.  476.65129. 

4.  47.86083. 


5.  594.691 

6.  112,7':  24. 

7.  223.M07744. 


8.  1380.0648193. 

9.  5974.10371. 
10.  339.626177443 


SUBTRACTION  OF  CIRCULATING  DECIMALS.— ART.  363. 


1,  2.  Given. 

3.  391.5524. 

4.  3.81824. 


5.  4.789. 
C.  400.915. 
7.  3.9046. 


8.  218.60. 

9.  0.013640731. 
30.  2451.386. 


PAGES  225— 235.] 


ANSWERS. 


413 


MULTIPLICATION  OF  CIRCULATING  DECIMALS.— ART.  3C  I. 


Ev       ANS. 

Ex.       Ass 

Ex.      ANS. 

1,  2.  Given. 
3.  0.082. 
4.  1.8. 

5.  380.185. 
C.  778.14. 
7.  750730.518. 

S.  31.701. 
0.  34008.4199003. 
10.  2.297. 

DIVISION  OF  CIRCULATING  DECIMALS.— ART.  3G5. 


1,  2.   Given. 

3.  55. GO. 

4.  5.41403. 


5.  7.72. 

C.  8574.3. 

7.  3.50G493. 

8.  3.145. 


0.  62.323834196- 

soi. 

10.   1.4229240011- 
85770750088. 


ADDITION   OF   FEDERAL   MONEY.— ART. 


1.  Given. 
2.  8205.04. 
3.  £581.128. 
4.  £500.50. 
5.  £1705.34. 
G.  £1431.50. 

7.  £3531.432. 
8.  £12200.524 
0.   £185.285. 
10.  £74.33. 
11.   £350.32. 
12.  £0401.05. 

13.   £8705.12. 
14.  £10080. 
15.   £378.383. 
10.   £300.100. 
17.  £250.213. 

18.  £1045.258 
10.  £821  10.17. 
20.  £71774.75. 
21.  £27800.74. 
22.  £81800.03. 

SUBTRACTION  OF  FEDERAL  MONEY.—  ART.  375. 

1.   Given. 
2.  £12.13. 
3.  £84.82. 
4.  £247.15. 
5.  £018.48. 

0.  £183.22. 
7.  £323.47. 
8.   £373.82. 
0.  £10870.75. 
10.  £1000.49. 

11.  $0047.788. 
12.  £011  10  304 
13.  £18.081. 
14.  £88.11. 
15.  £180.02. 

10.   £2.037. 
17.  £3  2.  M50. 
18.   £10800.07. 
10.  £80080.90. 

MULTIPLICATION  OF  FEDERAL  MONEY.—  ARTS.  377,  378. 

3.  £83.00. 
4.  £517.025. 
6.  £30.50375. 
7.  £1440.75. 
8.  £40.50375. 

0.  £84.875. 
10.  £103.75. 
1  I.  £205.025. 
12.  £320.25. 
13.  £2.0025. 

14.  £2.84375. 
15.  £000.375. 
10.  £2.70. 
17.  £14.0025. 
18.  £15.78375. 

19.  £28.125. 
20.  £220.50. 
21.  £142.50. 
22.  £2331.875. 
23.  £14084.  12f 

DIVISION  OF  FEDERAL  MONEY.—  ARTS.  379-3§I. 

1.  Given.                       5.   Given.                     10.  543.518  +  yds. 
2.  £4.50.                       6.  8.207  coats.            11.  001.421+doz. 
3.  £0.00.                       7.   7.871+  times.       12.  300  skeins. 
4.  $3.13.                       9.  308.035+  gals.     13.  $3.524  +  . 

114 


ANSWERS.  [PAGES  230 — 248 


DIVISION    OF    FEDERAL    MONEY    CONTINUED. ART.   3§1. 


Ex.                AN* 

fi\.                        A  MS. 

Ex.               Ass. 

14.  81.50. 
15.  80.25. 
IG.  81.073  +  . 
17.  83.015+. 
18.  80.084  +  . 

10.  80.04040  +  . 
'20.  80.02700  +  . 
21.  8J.78008  +  . 
22.  81.5435  +  . 
23.   1714.285+  bu. 

24.   113.50377+  tons 
25.  80.505238  +  . 
20.  245.517+  acres. 
27.  500  cows. 
28.   150  carriages. 

APPLICATIONS  OF  FEDERAL  MONEY.—  ARTS.  3§2-§5. 


1.  Given. 

13.  $5885. 

28.  $0.0072. 

2.  8800. 

14.  $10538.625. 

20.  $0.0064. 

3.  8511.50. 

15,   16.   Given. 

30.  $13.4710  + 

4.  8780. 

17.  86.33375. 

per  cwt.1; 

5.  8780. 

18.   $104.55. 

$0.134710  + 

6,  $1350. 

10.   $114.108. 

per  Ib. 

7.  $1020. 

20.  $50.5856.               \3l.  $12.88506  cwt. 

8.  $864.50. 

21.  $505.3775. 

$0.1288506  Ib. 

9.  82418. 

22.  $1001.75. 

32.  $120.625. 

10.  $4440. 

23.  $5.40625. 

33.  $208.838. 

11.  $1424.75. 

24.  $52.126. 

34.  $1734.875. 

12.  $2691.875. 

25.  $437.645. 

35.  $13703.78. 

PERCENTAGE.—  ART.   388. 


5,  6.  Given. 

7.  $7.6875. 
8.  $8.7526. 
9.  $3.4608. 
10.  $8.7078. 
11.  $114.1070. 
12.  $10.50. 
13.  $210. 

14.  $43.13  rec'd. 
$810.43  paid. 
15.  $402.05. 
16.  $134. 
17.  $32.625. 
18.  $34.03575. 
19.  $62.50. 
20.  $146.666+. 
21.  $8.771875. 

22.  375  sheep. 
23.  $1568. 
24.  187.5  lost  ; 
1312.5  saved. 
25.  $8.125. 
26.  $6.316. 
27.  $84.52016. 
28.  $250. 
29.  $750. 

30.  $90.4824. 
31.  $844.08. 
32.  $4724.775. 
33.  $1250. 
34.  $12000. 
35.  $21900,  1st; 
$14600,  2d 
36.  $200. 
37.  $0.95. 

APPLICATIONS    OF   PERCENTAGE.— ARTS.  395-97. 


1.  Given. 
2.  $12.507. 
3.  $58878. 
4.  $73.159. 
5.  $115.203. 
6.  $615. 
7.  $583.842. 
8.  $52.834. 
9.  $155.875. 

10.  $619.887. 
11.  $44.32. 
12.  $673.75. 
13.  $57.29. 
14.  $415.831. 
15.  $106.831  A. 
$2029.7080. 
17.  $21078.431. 
18.  $3439.613. 

19.  $761904.761. 
20.  $4126.55. 
21.  $1413.975. 
22.  $46.50. 
23.  $22.113. 
24.  $9.375. 
25.  $318.975. 
28.  $3692.50. 
29.  $2250. 

30.  $8840.70. 
31.  $7072. 
32.  $3552.50. 
33.  $1350  rec'd 
$180  lost. 
34.  $7490.50. 
35.  $960. 
36.  $4427.50. 
37.  $9028.50, 

PAGES  251 270.]  ANSWERS. 


415 


INTEREST.— ART.  4O4. 


Ex.            ANS. 

Kx.            A  S3 

Kx              ANS. 

Ex             A  us 

1.   S'-29.B  1. 
2.  $43.255. 
3.  $40.367. 
4.  $51.20. 
5.  $60.263. 
6.  $44.414. 
7.  $194.58. 
8.  $17.803. 
9.  $28.206. 

10.  $8.103. 
11.  $6.853. 
12.  $19.14. 
13.  $60.27. 
14.  $89.40. 
15.  $958.41. 
16.  $657.45. 
17.  $1006.833. 
18.  $1585.018. 

19.  $889.44. 
20.  $1.135. 
21.  $1.409. 
22.  $1.898. 
23.  $102.125. 
24.  $154.216. 
25.  $704.083. 
26.  $2.975. 
27.  $76.131. 

28.  $3312.209. 
29.  $5278.162. 
30.  $16.158. 
31.  $206.718,  at 
360  days  ; 
$203.886,  at 
365  days. 
32.  $66778.64 

SECOND   METHOD.— ARTS.  409—413. 


4.  $8.50. 

15.  $82.078. 

26.  $307.65. 

38.   $137.288. 

5.  $1.065. 

16.  $39.179. 

27.  $227.994. 

39.  $481.016. 

6.  $70.151. 

17.  $320.833. 

28.  $8. 

40.  $391.062. 

7.  $97.28. 

18.  $9.8437. 

29.  $0.07. 

41.  $1531.25. 

8.  $30.78. 

19.  $85.207. 

31.  $15.60. 

42.  $3425.655. 

9.  $398.287. 

20.  $400. 

32.  $21.09. 

43.  $16320.528. 

10.  $1177.50. 

21.  $1638.442. 

33.  $1.272. 

45.  $2.145. 

11.  $1113.024. 

22.  $144. 

34.  $4.778. 

46.  $74.392. 

12.  $10.05. 

23.  $90. 

35.  $46.35. 

47.  $10.835. 

13.  $11.0025. 

24.  $12666.075. 

36.  $129.15. 

48.  $398.055. 

14.  $988.761. 

25.  $16360.996. 

37.  $168.552. 

49.  $14.532. 

APPLICATIONS   OF  INTEREST.—  ARTS.  415—410. 

2.  $5.25.               7.  $36.08. 

12.  $6547.20. 

20.  £8,  18s.  9  d. 

3.  $3.15. 

8.  $91.085. 

14.  $499.034. 

21.  £12,  10s. 

4.  $17. 

9.  $107.854+. 

15.  $498.595. 

22.  £1898,  10s. 

5.  $60. 

10.  $533.867. 

16.  $4149.689. 

4|d. 

6.  $45.014. 

11.  $25729.166+ 

19.  £19,  5s.  10£d23.  £2900. 

PROBLEMS   IN   INTEREST.—  ARTS.  422-424. 

1,  2.  Given. 

10.  5  per  cent. 

19.  $30000. 

28.  14  y.  3  mo. 

3.  6  per  cent. 

11.  5"  per  cent. 

20.  $20833^. 

13d.  nearly. 

4.  6  per  cent. 

13.  $1800. 

22.  4  years. 

29.  14  y.  3  mo. 

5.  8  per  cent. 

14.  $5400. 

23.  6  months. 

13d.  nearly. 

6.  7£  per  cent 

15.  $10000. 

24.  1  y.  3  mos. 

30.  10  years. 

7.  5$  per  cent. 

16.  $8000. 

1  d.  nearly. 

31.  8  y.  4  mo. 

8.  7  per  cent. 

17.  $14285.7143. 

25.  1  y.  6  mo. 

32    9  y.  6  mo.  8  d 

9.  6  per  cent. 

18.  $20000. 

27.  16  y.  8  mo.     S3.  28  years. 

COMPOUND   INTEREST.—  ARTS.  426,   t27. 

1,  2.  Given. 

5.  $4590.09. 

8.  $1551.328. 

11.  $16035.675. 

3.  $507.213. 

6.  Given. 

9.  $877.506. 

12    $149744. 

4.  $2177.426. 

7.  $1888.464. 

10.  $3491.395. 

416 


ANSWERS. 


P  AG  ES  272  -  295 


DISCOUNT.— ART.  43O. 


Ex.            A\s 

Ex.             ANS. 

Kx.             ANS. 

Ex.            AN* 

'1,  "2.  Given. 
?,.  $934.579-f. 
4.  $1488.087+. 

5.  $88.461+. 
6.  $83.52+. 
7.  $4729.064+. 

8.  $6208.955+. 
9.  $3404.347-f  -• 

10.  $Wo«.2l8+. 

11.  $36.t'3t>. 

BANK  DISCOUNT.—  ARTS.  433,  431. 

12,  13.  Given. 

20.  $24.822. 

27.  $456.785. 

34.  $3821.883. 

14.  $14.18-25. 

21.  $48.3237. 

28.  $1126.523. 

35.  $4355.102. 

15.  $1(5.605. 

22.  $37.595. 

29.  Given. 

36.  $63717.884. 

16.  $26.98. 

23.  $43.694. 

30.  $414.507. 

37.  $10416.666. 

17.  $5.495. 

24.  $6381.59. 

31.  $966.101. 

38.  $51194.539. 

18.  $2034.1213. 

25.  $1495.625. 

32.  $1252.70. 

39.  $46638.655. 

19.  $2774.655. 

26.  $80. 

33.  $2514.247. 

40.  $8301.342. 

INSURANCE.—  ARTS.  437—442. 

1.  Given. 

8.  $1875. 

16.  1  per  cent. 

25.  $8365.482. 

2.  $20.70. 

9.  $487.50. 

17.  H  per  cent. 

26.  $13876.288. 

3.  $94.20. 

10.  $243.125. 

19.  $52000. 

27.  $27027.027 

4.  $63.75. 

11.  $192.78. 

20.  $65600. 

29.  $48.60. 

6.  $104. 

12.  $3375. 

21.  $65000. 

30.  $373.75. 

6.  $70.50. 

14.  2i  per  cent. 

22.  $573331. 

31.  $1UOOO,  ins. 

7.  $900. 

15.  2£  per  cent. 

23.  $3416s. 

$12250,prem 

PROFIT  AND   LOSS.—  ARTS.  444—447. 

1-3.  Given. 

10,  11.  Given. 

19.  15|  per  cent 

27.  $2622.222. 

4.  $218. 

12.  $156.804. 

20.  100  per  cent. 

28.  $2-736. 

5.  $680. 

13.  $4238.50. 

21.  20|H  perct 

.29.  $13043.478. 

6.  $935.25. 

14.  $5926.85. 

22.  2f  •»•  per  cent 

30.  $6317.391. 

7.  $1366.75. 

15.  $29504.875. 

23,  24.  Given. 

31.  $17806.122, 

8.  $68730.28. 

17.  23,-^-  per  ct 

25.  $460.869. 

32.  $42654.028. 

9.  $12500  lost. 

18.  4£  per  cent. 

26.  $205.882. 

33.  $42160. 

DUTIES.—  ARTS.  451-453. 

1.  Given. 

7.  $3784. 

13.  $717.40. 

19.  $12642.40. 

2.  $370.80, 

8.  $345.744. 

14.  $492. 

20.  $2807.10. 

3.  $163.20. 

9.  $679.14. 

15.  $1051.71. 

21.  $11172.30 

4.  $1323. 

10.  $1882.406. 

16.  $715.75. 

22.  $17328.75. 

5    $546. 

11.  Given. 

17.  $1230. 

23.  $15770.70. 

6.  $1235.22. 

12.  $248. 

18.  $15884.75. 

ASSESSMENT  OF   TAXES.—  ARTS.  456,  457                   . 

1,  2.  Given.                     5.  f  of  1  per  cent.,  or     7.  $121.  PL,  ITs  (ax. 

3.  $54.15,  B's  tax.         .      8  mills  on  $1.             8.  $283.68,  C's  tax. 

4.  $80.50,  C's  tax.          6.  $80,  A's  tax.             10.  $8854.166. 

PAGES  296 310.J  ANSWERS. 


417 


ASSESSMENT  OF  TAXES  CONTINUED. ARTS.  459,    460. 


Ex                 ANS 

Ex.                Ass.                       |  Kx.                ANS 

11.  $  161  25.654. 
1-2.  Si  7342.  105. 
13.  .$34051.815. 
16.  $73,  G.  A's. 
17.  $116,  H.  B's. 
18.  $451.50,  W.  C's. 
19.  $481  22,  E.  D's. 

20.  $314.50,  J.  F's. 
21.  $621.9J,  T.  G's. 
22.  $526.40,  W.  H's. 
23.  $263.30,  L.  J's. 
24.  $631.00,  W.  L's. 
25.  $196.90,  J.  K's. 
26.  $404.90,  G.  L's. 

27.  $370.50,  F.  M'a. 
28.  $458.20,  C.  P's. 
29.  $480.50,  J.  S'a. 
30.  $541,  R.  W's. 
32.  $13.36. 
33.  $3.45. 
34.  $13.40. 

ANALYSIS.— ARTS.  462— 47O 


1, 

2.  Given. 

12. 

$0.039-tV. 

22.  $7.98. 

32. 

Given 

3. 

$300. 

13. 

$04. 

23.  $6.03. 

33. 

300  Ibs. 

4. 

$320. 

14. 

$1080. 

24.  $160. 

34. 

1500  Ibs. 

5. 

$12.33!. 

15. 

$480. 

25.  $3.70?. 

35. 

95.2  cords. 

6. 

$10.50. 

10. 

$8. 

26.  $3430. 

30. 

100  pair. 

7. 

$1.08*. 

17. 

GO  days. 

27.  $119.918. 

38. 

$450, 

A  s. 

8. 

$2640. 

18. 

29!f 

mos. 

28.  $636.479. 

$750, 

B's. 

9. 

$24.80. 

19. 

1088 

days. 

29.  Given. 

39. 

$450, 

A's. 

10. 

$0.055. 

20. 

$0.50 

. 

30.  2£  hours. 

$000, 

B's. 

11. 

$0.29|. 

21. 

$3. 

31.  2!£  days. 

$750, 

C's. 

40. 

$763.  03-^-, 

A's. 

49. 

40  tons.  A's. 

65. 

188! 

Ibs.  at  8d. 

$054.54-^1-, 

B's. 

80  tons,  B's. 

17! 

Ibs.  " 

12d 

$981.81-^, 

C's. 

120  tons,  C's. 

in 

Ibs.  « 

18d, 

41. 

$150.  95f(f-f, 

A's 

50. 

25  per  cent. 

17! 

Ibs.  " 

22d, 

$164.53fH 

B's 

51. 

33!  Per  cent. 

67. 

9'  horses. 

$185.70/1^ 

C's 

$30000,  loss. 

68. 

38|  days. 

$123.803V3 

D's 

53. 

5s.  per  gal. 

69. 

278160  men: 

42. 

6G|  cts.  on  $1. 

54. 

5fs.  per  Ib. 

70. 

15!  7  months 

$260.  GOf,  1st. 

55. 

9  cts. 

per  Ib. 

71. 

$54.60. 

$333.33|,  2d. 

50. 

19!  cts.  per  Ib. 

72. 

$459. 

90, 

$400.000,  3'd. 

57. 

91^  cts.  per  gal. 

74. 

$600. 

43. 

70  cts.  on  $1. 

58- 

60.  Given. 

75. 

$3600 

44. 

25  per  cent. 

01. 

1  part 

16  car. 

76. 

$030. 

45. 

$2990.00,  A's. 

1     " 

18  car. 

77. 

72. 

$4197.50,  B 

's. 

2!  " 

23  car. 

78. 

300. 

$4312.50,  C's. 

r  " 

24  car. 

79. 

120. 

46 

OGf  per  cent. 

03. 

lOOgals.  at80cts. 

80. 

240. 

47 

37!  Per  cent. 

40 

"       30cts. 

81. 

G 

8!f 

48 

10  per  cent. 

40 

"      40cls. 

85. 

$230. 

418 


ANSWERS.  [PAGES  311 — 328 


ANALYSIS    CONTINUED. ART.    471. 


Ex.     ANS. 

Ex.     ANS. 

Ex.     ANS. 

Ex.     APS. 

86.  $1170. 

100.  £266. 

114.  $288. 

127.  $140. 

87.  $900. 

101.  £131| 

115.  $43|. 

128.  $1560. 

88.  $1125. 

102.  £353. 

116.  $814. 

129.  $180. 

89.  1367.50 

105.  $2501. 

117.  $640. 

130.  $630. 

90.  $442. 

106.  $231. 

118.  $3000. 

131.  $180. 

91.  $201. 

107.  S119-&. 

119.  $200. 

132.  $1281. 

92.  $350. 

108.  $186. 

121.  $625. 

133.  $800. 

93.  $240. 

109.  $280.    i!22.  $480. 

134.  $12.60. 

94.  $754. 

110.  $1170.   ;123.  $808. 

135.  $45. 

95.  $1080. 

111.  Given. 

124.  $420. 

136.  $45. 

96.  $630. 

112.  $378. 

125.  $690. 

137.  $90. 

<*9.  £206$. 

113.  $810. 

126.  $877£. 

138.  $150. 

RATIO.— ARTS.  48O— 48§. 


1    2.  Given. 

14.  8-H. 

26.   120. 

40.  45  to  72. 

3.  2. 

15-  i. 

27.  60. 

41.  Equal. 

4.  4. 

16.  i. 

28.  -fr. 

42.  936  to  56C. 

5.  9. 

17.  *. 

29.  V\i> 

43.  G.  inequality 

6.  6. 
7.  6. 

18.  i. 
19.  4. 

30.   1J. 
31.  i. 

44.  L.  inequality 
45.  Equality. 

8.  8. 

20.  |. 

32.  240. 

46.  60  :  12=5. 

9.  9. 

21.  3. 

35.  8;  f 

47.  1. 

10.  9. 

22.  7. 

36.  4;  8. 

48.  f. 

11.  9. 

23.  112  avoir. 

37.  44  \. 

49.  H. 

12.  9. 

24.  4. 

38.  *;  9. 

50.  ttfr- 

13.  4 

25.  6. 

39.  72  to  8. 

51.  f. 

SIMPLE  PROPORTION.—  ARTS.  5O2—  5O6. 

1.  12. 

17.  51ilbs. 

35.   1925}lbs.cop.j47.  480. 

2.  3. 

18.  $1640.64. 

641  |lbs.  tin. 

48.  375  sheep. 

3.  16. 

19.  $7066.40. 

36.  1520  Ibs.n. 

49.  20  days. 

4.  3. 

22.  3  far. 

280  Ibs.  c.      - 

50.  400  rods. 

5.  Given, 

23.  $792. 

200  Ibs.  sul. 

51.  8-f  weeks. 

6.  20. 

24.  $2768. 

37    980.5155  Ibs. 

52.  £1,    3s.    6d 

7.  55*. 

25.  435  miles. 

38.  $1350. 

1  iV  for,  1st 

8    120. 

26.  252  days. 

39.  £45. 

£1,    Is.    2d 

9,  10.  Given 

28.  $2.70. 

40.  $3375. 

-ft-  far.  2nd. 

11.  $903. 

29.  3s.  3d.  2£fq. 

41.  $2562.50. 

£0,   18.'    9d. 

12.  $1309.50 

30.  $3.15. 

42.  $16480. 

3r\fai.3rd. 

13.  $-225. 

31.  $8.555. 

43.  70400  times. 

£0,   16s.  5d. 

14.  775  miles. 

32.  $26.40. 

44.  57600  imes. 

2U  far.  4th. 

15.  20  tons. 

34.  30   bu.  oats; 

45.   170. 

53.  888  4  oz.  ox. 

16.  2156  Ibs. 

70  bu.  corn. 

46.  200                 -1        llli  oz.  hy. 

PAGES  329 355.]  ANSWERS. 


419 


COMPOUND   PROPORTION. — ARTS.  5O§— 51 1. 


Ex.            AN*. 

Ex.            ANS. 

Ex.            ANS. 

Ex.            ANS. 

3.   10  horses. 
4.    19*-  days. 
5    1314  gals. 
6    27  laborers. 

9.  24  days. 
10.  144  days. 
11.   1125  miles. 
12.  $225. 

13.  $140. 
14.  $768. 
15.  $600. 
16.  32  days. 

17,  18.  Given. 
19.  56  yds.  Can. 
20.  127  b.  N.  O. 
21.  16  rupees. 

1,  2.  Given. 

3    28  sq.  ft.  6'  10". 

4.  59  cu.  ft.  3'  8". 

5.  268  cu.  ft.  6'  11" 

6.  235  sq.  ft. 

7.  734  sq.  ft.  0'  9". 


DUODECIMALS.— ART.  516. 

8.  105ft.  5' 4"  5'"  5""  1 11.  195  ft.  4' 1"  3"  3" 

4 0 6""". 

9.  154 ft.  3'  1"  5'"  4"" 1 12.  23  C.  Ill  ft.  3' 

6'""  8""".  !13.  3840  ft.  0'  5". 

10.  85ft.  I'll"  0'"  5""  14.  $15.819-|-. 


2'""  6""". 


15.  33750  bricks 


EQUATION   OF    PAYMENTS.— ART.  521. 
3.  6  months.   *    |  4.  6  months.       |  5.  3  years.          |  6.  62  days. 


PARTNERSHIP.— ART. 


1.  Given. 

2.  $240,  A's  gain. 
$320,  B's  gain. 
$400,  C's  gain. 

3.  $274.21  f^,  A's. 
&373.40W,  B's. 
$212.37-Hi,  C's. 

4.  $1178.947,  A's. 


$2259.649.  B's. 

$3340.351,  C's. 

$4421.053,  D's. 
5.  $850,  A's. 

$800,  B's. 

$700,  C's. 

$650,  D's. 
7.  $1655.172,  X's. 


523. 

$1448.276,  Y's. 
$1396.552,  Z's. 

8.  $22.486,  A's. 
$21.024,  B's. 
$16.490,  C's. 

9.  $3492.06,  A's. 
$4761.91,  B's. 
$6746.03,  C's. 


EXCHANGE  OF  CURRENCIES.— ARTS.  533—537. 


3.  $4116.42. 

4.  $850.63. 

5.  $414.667. 

6.  $969.815. 

7.  $2041.59-)-. 

8.  $4841.089-|-. 

9.  $7746.082+. 

10.  $60652.55-}-. 

11.  $208683.8 19+ 
32    $330661.6054-. 

13.  $242840.3694- 

14.  $257791.3971-. 


2.  $4791.60. 

3  $25391.084+. 

4  $284.58. 


15.  $369716.864+. 

16.  $284412.622+. 

17.  $4840000. 

19.  £82. 

20.  £90. 

21.  £181.  lOid. 

22.  £261.  8s.  7£d. 

23.  £446,  7s.  8i,d. 

24.  £201,  11s.  7 |d. 

25.  £883,  5s.  8Jd. 

26.  £1095,  3s.  ll|d. 

27.  £5220,  9|d. 


28.  £8568,  3s.  7}d. 

29.  £10384,  18s.  4d. 

30.  £20661,  3s.  lid. 

32.  £135. 

33.  £227. 

34.  £315,  9d. 

35.  £375. 

37.  $534.166. 

38.  $614.1875. 

39.  $986.083. 

40.  $7714.285. 

41.  $20000. 


EXCHANGE.— ART.  548. 

5.  $10152-.527+. 

6.  $707. 

7.  $1881.60. 


8.  $15418.509. 

9.  $20665.20. 
10.  $36.480.755. 


420 


ANSWERS.  [PAGES  356 — 378 


ARBITRATION   OF   EXCHANGE.— ART.  519. 


Ex.               Ana. 

Ex. 

ANS. 

Ex.               ANS. 

1.  21  florins. 

2.  $ 

45  gain. 

3.  180  milrees,  circu 

ALLIGATION.— ARTS.  552—556. 


2.  $0.87i. 

3.  5s.  4d.  l±-§  qr. 

5    3  grs.  at  1 8  car.  fine. 
1  gr.  "  20       " 
1  gr.  "   22       " 
&  grs."   24       " 


1.   10  oz.  16  car.  fine. 
5  oz.  18 
5  oz.  22       " 

8.  133  Ibs.  at  20  cts. 
95  Ibs.  at  30  cts. 
190  Ibs.  at  54  cts. 


10.  40  gals,  at  15s. 
40  gals,  at  17s. 
40  gals,  at  18s. 
200  gals,  at  22 
28  gals,  water ; 
98  gals.  wine. 


11 


INVOLUTION.— ART.  562. 


17.  3125. 

22.  6.25. 

27.  VVW- 

18.  279936. 

23.   .000001728. 

•28.  -rfjj-JH 

19.  117649. 

24.  .0000015625. 

29.  20}. 

20.  65536. 

25.  t-. 

30.  54ff-. 

21.  387420489. 

26.  T^. 

31.  1480-f 

IS.  Given. 
13.  15129, 
14  2460375. 

15.  8294400. 

16.  10000. 


3.  51. 

4.  73. 

5.  28. 

6.  9.327-J-. 

7.  69. 

8.  84. 

9.  99. 

10.  167. 

11.  31. 

APPLICATIONS  OF  THE  SQUARE  ROOT.— ARTS.  581-585. 


SQUARE  ROOT.—  ARTS.  574,  575. 

12.  9.848+. 

21.  792. 

30.  186.9951-f 

13.  2.6457+. 

22.   1.7810+ 

31.  12345. 

J4.   13.78404+. 

23.  3216. 

32.  345761. 

15.  209. 

24.   $-. 

33.  31.05671. 

16.  217. 

25.  f-V. 

34.  19.104973174 

17.  23.8. 

26.  .79056+. 

35.  1.41421356- 

18.  2.71. 

27.  4.1  683+. 

237  + 

19.  .9044+. 

•28.  28.181. 

36.   1.732050807 

20.  34.2. 

29.  14.4116+. 

5688772. 

1.  Given. 
2.  32  feet. 
3.  166.709+m. 
4.  240rds.  side. 
339.41  12  r.d. 
5.  Given. 
6.  10. 

7.  18. 
8.  36. 
9.  40. 
10.  66. 
11.   168. 
12.  11.2. 
13.  67.5. 

14.  ^ 

15.  n- 

16.  -ft**- 

17.  -Mr- 

18.  63  rods. 

19.  160  rods. 


20.  320  rods. 

21.  480,  length; 
160,  breadth. 
148  in  rank; 

74  in  file. 
25  and  40. 
18  and  47. 


22. 


EXTRACTION  OF  THE  CUBE  ROOT.— ARTS.  590-92. 


4  45. 

5  52. 

6.  83. 

7.  136. 

8.  217. 
9  22.6 

10.  2.74. 


11.  0.623. 
12.  3.332222-4-. 
13.  1.817121+. 
14.  7.  21  7652+. 
15.  8.315517+.  x 
16.  Jf-. 
17.  -U. 

18.  3.5463+, 
19  3|. 
20.  1.25992104. 
21.  .64:5(55958974. 
22.  68  ft. 
24.  3.1748+yds. 
25.  4  Ibs. 

27.  24  and  72. 
28.  128  and  256. 
29.  60  and  cOO. 
30.  160  and  640, 
31.  426  and  2556. 
32.  74  7  and  6723. 

f  AGES  379—393.] 


A  N  S  WE  RS. 


421 


ROOTS  OF  HIGHER  ORDERS.— ARTS.  593-i>5. 


E».        ANrf. 

Ex.         Ass. 

Ex.         AN*. 

Ex.        Ass. 

Cx. 

ANS. 

2.  2. 
3,  16. 
4    378. 

5.  6. 
6.  26. 
7.  5. 

8.   7. 

9.  3. 
10.  2. 

12.  2.4872+. 
13.  41  4.5+. 
15.   1.104089. 

16. 
17. 
18. 

1.080059. 
1.004074. 
1.047128. 

ARITHMETICAL   PROGRESSION.— ARTS.  6O3— 60S. 


1.  Given. 

2.  5050. 

3.  78  strokes. 


4.  Given. 

5.  33. 
7.  44. 


9.  3f. 

11.  33}. 

12.  502. 


13.  14,  21.  &  28 

14.  15,29,43,57, 

71,  &  85, 


GEOMETRICAL   PROGRESSION.— ARTS,  G1O-12. 


2.  4. 

3.  4374. 

4.  13671875. 

5.  $2048. 

6.  $334.5563944,  amt. 

of  $250. 


$750.3651759245, 
amt.  of  $500. 

$1628.894614622- 
37890625,  amt. 
of  $1000. 


8.  1023 

9.  43774}. 

10.  $111111111.111, 
12.  li. 
14.  3. 


1,  2.  Given. 
3,  $826.992. 


ANNUITIES.— ARTS.  614,  615. 

4.  $2298.262.     I  6.  36785.59.       I  8.  $1333.333 

5.  $4835.74.          7.  Given.  9.  Gi/-?n. 


PERMUTATIONS  AND  COMBINATIONS.— ARTS.  61  §,  619. 

2.  40320  ways.  |  4.  3628800  ways.         |  7.  15120  numbers. 


3.  362880  ways. 


5.  479001600  days.      |  8.  165765600  words. 


MENSURATION  OF  SURFACES.— ARTS.  622—631. 


1.  270  acres. 

2.  72:2  £  acres. 

3.  3H  acres. 

4.  320  rods,  or  1  m. 
6.  360  sq.  ft. 

6.  435  sq.  ft. 


7.  1100  sq.,ft. 

9.  290.4737  sq.  ft. 

10.  4  A.  52.82  rods. 

11.  62.8318  ft. 

12.  141.37155  rods. 


13.  100  ft. 

15.  12  A.  43.49375  r. 

16.  31415.9  sq.  ft. 

17.  2  ft.  9.94  in. 

18.  17.3205  ft. 


MENSURATION  OF  SOLIDS.— ARTS.  633—647. 


1.  1364  cu.  ft. 
2.  3154  ft.  11'  6"  8"'. 
3.  2615  cu.ft.  1080  in. 
4.  115  ft.  114.368  in. 
5    53333  J  cu.  ft. 
C.   8835.75.  cu.  ft. 
7.  900  sq.  ft. 
8.   1739  sq.  ft. 
9.  76  cu.  ft. 
10.  176  sq.  ft. 

11.  2748.891  25  cu.ft. 
12.  11  9366.25  cu.  ft. 
13.  78  vds.    4  ft.  123- 
.11  28  in. 
14.  7  sq.  ft.  9.87516  in. 
15.  14684558.20796 
sq.  miles. 
16.  1767.14437  cu.  in. 
17.  5291335807.60158 
cu.  in. 

18.  13  sq.  ft. 
19.  4  cu.  ft. 
20.  624  cu.  ft. 
21.  220  gals.  3qts.  1  pt 

22.  451    gals.    2    qta. 
0.729344  pt. 
23.  »31.71f»26-f  tons. 
24.  967.105214  ton*. 

422 


ANSWERS.  [PAGES  394 — 398 


MECHANICAL   POWERS.— ARTS.  618—655. 


Ex.            ANS. 

Ex.             ANS. 

Ex.            ANS. 

Ex.           ANS. 

1.  500  Ibs. 
2    133^  Ibs. 
*.  96  Ibs.  A.; 

160  Ibs.  B. 
4.  4  ft.  from  A.  ; 
8  ft.  from  B. 

5.  600  Ibs. 
6.  10661  Ibs. 
7.  1600  Ibs. 

8.  1250  Ibs. 
9.  11  36.3636  Ib. 
10.  904777.92  Ib 

MISCELLANEOUS   EXAMPLES. 


459  less. 

34.  136  g.  1  q. 

$440,  C's  g. 

79,  247170562- 

521  greater 

35.  $180. 

$700,  B's  s. 

2710s.  m. 

70. 

36.  $10.875. 

$1100,C'ss. 

80.  33600914- 

5^r. 

£7.  "$156.615. 

59.  20  per  cent. 

2264006.2- 

Nfc 

38.  94    d.  3  h. 

60.  $1371. 

3104  c.  m. 

20  days. 

38m.  10-Hs. 

61.  $4755.141. 

81.  5890.5  Ibs. 

$61.32. 

39.  $2. 

62.  $32000. 

82.   585.80357b 

$16581.65. 

40.  £1. 

63.   $360. 

83.  39.401  hhd. 

$18.60. 

41.  *|f. 

64.  36  days. 

84.  7ift. 

$1843.003. 

42.   $4800 

65.  90  hours. 

85.  403291461- 

$24390.243 

43.  $197.759. 

66.  £51,  A's. 

126605635- 

$4.50. 

44.  228  gals. 

£34.  B's 

584000000. 

$6.875. 

45.  $40.29$. 

£68,  C's 

86.  31m.  180  r 

33i  per  ct. 

46.  $41.095. 

£102,  D's. 

87.  662^. 

$36. 

47.  2y.  182-J-d. 

67.  £160,  A's. 

88.  $4294967- 

$229|f. 

48.  5-fr  min. 

£224,  B's. 

.295. 

4987i|E. 

49.  120  days 

£256,  C's. 

89.  5  bags,  A. 

2000  miles. 

50.   120  schol. 

£205,  A's. 

7  bags,  M. 

240*0  times. 

51.  £292. 

£287,  B's. 

90.   1440. 

2880  times. 

52.  $6000. 

£328,  C's. 

91.  $230,  B's. 

$1.50perg. 

53.   600. 

68.  $520,  D's. 

$325,  C's. 

2i|  cts. 

54.  5600  Ibs.  t. 

$280,  A's. 

$445,  A's. 

$2400. 

750  Ibs.  1. 

$360,  B's. 

92.  5    o'clock, 

437|  bbls. 

300  Ibs.  b. 

69.  20. 

20  min. 

21  months. 

55.  254|  miles. 

70.  25  persons. 

93.  lOffi  d.  all 

$1.328. 

56.  78f  Ibs. 

71.  40  and  80. 

47fi  d.  A 

40  yds. 

Il7f  Ibs. 

72.  75  and  128. 

38ff-  d.  B. 

Is.  3  gV  qrs. 

57.  $192.3071a3- 

73.  56.5685  ft. 

27-ftVd.C. 

37ffHfd< 

A's  gain. 

74.   7200  rods. 

111^  d.  D. 

34*W*d. 

$2307.692 

75.  3.535519ft. 

94.   36£  days. 

$1.60.. 

•i^B'sg. 

76.  677.-73475f. 

95.   12    o'clock. 

12  miles. 

$2500.000, 

77.  7.13645  r. 

32-^  min. 

12f  days. 

C'sg. 

78.  50  A.  3  R. 

96.  128tyrs. 

52£  days. 

58.  $240,A'sg. 

28.7399+r. 

97.  $407. 

•r 


,' 


